Thermal Properties of Matter Aakash RM Modules (@TEAMFLOOD) - Unlocked

Chapter 11
Thermal Properties of Matter

Chapter Contents
z Temperature and Heat TEMPERATURE AND HEAT
z Measurement of Temperature is a relative measure, or indication of hotness or coldness.
Temperature
The SI unit of temperature is kelvin (K), whereas degree celsius (°C) is a
z Thermal Expansion commonly used unit of temperature. Energy transfer that takes place solely
because of a temperature difference is called heat. The SI unit of heat energy
z Specific Heat Capacity transferred is expressed in joule (J). In conventional system, the unit of heat is
z Calorimetry called calorie (cal) and 1 cal = 4.186J.
z Change of State A hot body has more internal energy than another identical cold body.
z Heat Transfer MEASUREMENT OF TEMPERATURE

z Newton’s Law of Cooling The instrument used to measure temperature called thermometer. To construct
a thermometer, we use any thermometric property of any substance which
z Black Body Spectrum
varies proportional to temperature.
z Some Important Formulae To measure the temperature we assume to fixed temperature i.e freezing point
and boiling point of water at atmospheric pressure and given them number
0 and 100 respectively.
The temperature difference between 0 and 100 divide in 100 equal parts each
part called 1 degree.
The value of degrees measured on different scales.
100°C 373 K 212°F
C K F
0°C 273 K 32°F

Celsius scal e Kelvin scale Fahrenheit scale
Measured temperature – LFP

 constant
UFP – LFP
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108 Thermal Properties of Matter NEET
C 0 K  273 F  32 Measured temprature – LFP

  
100  0 373  273 212  32 UFP – LFP
C K  273 F  32
 
5 5 9
Example 1 : What is that temperature at which the Celsius and Fahrenheit scale give the same temperature
value?
Solution : Let that temperature be x, then
C –0 F – 32
∵ 
100 – 0 212 – 32
Now, put C = F = x
x–0 x – 32
or, 
100 – 0 180
or, x = – 40
So, at – 40°C, Fahrenheit reading is also – 40°F.
Example 2 : What is that temperature at which the Fahrenheit reading is double that of the Celsius reading?
Solution : Let the Celsius reading be x then the Fahrenheit reading will be 2x.
x–0 2 x – 32
So, 
100 – 0 212 – 32
x 2 x – 32
or, 
5 9
or, 9x = 10x – 160
or, x = 160°C
So, 2x = 320°F.
Example 3 : Suppose that on a temperature scale X, water boils at –60°X and freezes at –180.5°X. What would
a temperature of 350 K be on the X-scale?
Solution : Upper fixed point = –60 °X

Lower fixed point = –180.5 °X
Suppose 350 K = x °X, then
x  ( 180.5) 350  273.15


 60  ( 180.5) 100
x  180.5 76.85
or,   0.7685
120.5 100
or, x = (120.5) (0.7685) – 180.5 = –87.9
 350 K = –87.9°X
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NEET Thermal Properties of Matter 109
z Different Types of Thermometers
1. Mercury Thermometer : Fahrenheit was the first to choose mercury as the thermometric substance
on account of its many advantages. It doesn’t wet glass, can be easily obtained pure, remains liquid
over a fairly wide range, has a low specific heat and high conductivity, it is opaque and its expansion
is approximately uniform and regular.
The range of an ordinary mercury thermometer is limited by the fact that mercury freezes at
– 38.87°C and boils at 356°C but the upper limit can be realised to about 500°C by filling the top of
the tube with nitrogen under pressure. If instead of mercury we use Alcohol, we can measure as low
as – 111°C, and petroleum ether can measure as low as – 190°C.
2. Constant Volume Gas Thermometers : Gas thermometers are more sensitive than liquid thermometers
as expansion of gases is more than that of liquids. They are most accurate thermometers. Their
temperature range is between – 260°C to 1600°C.
The pressure and temperature are related as Pt = P0(1 + t )
Pt – P0 t –0

P100 – P0 100 – 0
Pt  P0
t  100 C
P100  P0
3. Platinum Resistance Thermometer : The platinum resistance thermometer works on the principle of
Wheatstone bridge.
Rt = R0 [1 + t]
then the resistance and temperature are related as
Rt  R0 t 0

R100  R0 100  0
Rt  R0
t  100 C
R100  R0
These thermometers can measure temperature from – 182°C to 1200°C.
4. Thermoelectric Thermometers : It is based on Seebeck effect. In this thermometer two distinct

metals are joined to form a closed circuit known as thermocouple. One junction is generally kept in ice
and the other junction is placed on the body whose temperature is to be calculated. A current flows in
the circuit given by current, I = aT + bT 2 where a, b are constants and T is temperature. Such
thermometers have several advantages. They can easily measure fast changing temperatures. Since
temperature is in the form of electric current a computer can do the recording of changing temperature
over a time interval and it can even be sent through satellites etc. from one place to other. Different parts
of our body has different temperature and it can easily make temperature diagrams of our body to be
shown on a computer. It has very wide application in engineering and medical world. The temperature
range is (–200°C to 1600°C)
5. Pyrometers : These devices are based on Stefan’s law which says radiations received is proportional
to fourth power of the absolute temperature of the body. It measures very high temperatures [pyre means
fire] say that of a furnace etc. The temperature of the sun is measured by pyro-helio-meter.
6. Vapour Pressure Thermometers : Its use is based on the well-known fact that the vapour-pressure
of a liquid varies uniquely with the temperature. The method consists simply in measuring the vapour
pressure of a certain liquid at the required temperature.
7. Magnetic Thermometers : Temperatures below 1 K are measured by ‘magnetic thermometers’ which

uses the principle of adiabatic demagnetisation. It is based on Curie-Law according to which magnetic
susceptibility of a paramagnetic salt varies inversely proportional to its absolute temperature.
Example 4 : The pressures of the gas filled in the bulb of a constant-volume gas thermometer are 66 cm and
88 cm of mercury column at 0°C and 100°C respectively. When its bulb is immersed in a liquid
placed in a vessel, its pressure is 82.5 cm of mercury column. Calculate the temperature of the
liquid.
Solution : Pressure at 0°C is P0 = 66 cm, pressure at 100°C is P100 = 88 cm. Pressure at unknown
temperature t is Pt = 82.5 cm. According to formula, the unknown temperature is
 Pt  P0 
t = 100°C ×  P  P 

 100 0 
 82.5  66 
= 100°C ×  
 88  66 
16.5
= 100°C × = 75°C
22
Example 5 : The following observations were recorded on a platinum resistance thermometer : Resistance at
melting point of ice = 3.70 ohm, resistance at boiling point of water at normal pressure = 4.71 
and resistance at t°C = 5.29 ohm. Calculate
(i) Temperature coefficient of resistance of platinum,
(ii) Value of temperature t.
Solution : (i) Temperature coefficient of resistance is given by
R100  R0 4.71  3.70

 = R  100 =
0 3.70  100
1.01
= = 2.73 × 10–3 per °C
370
(ii) For temperature t, we have
R t  R0
t = 100°C × R  R
100 0
5.29  3.70 1.59

= 100°C × = 100°C × = 157.4°C
4.71  3.70 1.01
THERMAL EXPANSION
Most of the materials expands, when their temperature increases, with few exception such as carbon.
z Thermal Expansion and Interatomic Energy
We can understand thermal expansion qualitatively on the molecular level. Consider the interatomic forces in
a solid as springs, as in figure (a). There is an analogous relationship between spring forces and interatomic
forces. Each atom vibrates about its equilibrium position. When the temperature increases, the energy and
amplitude of the vibration also increase. The interatomic spring forces are not symmetrical about the equilibrium
position, they usually behave like a spring that is easier to stretch than to compress. As a result, when the
amplitude of vibration increases, the average distance between atoms also increases. As the atoms get farther
apart, every dimension increases.
U(r)
r = distance between atoms
Average distance between atoms
r0 = average distance between atoms
r0 rav
O r
rav
E3
E2
rmin E1
So solid expands
rmax on heating
As energy increase from
E1 to E2 to E 3, average distance
Fig.: (a) A model of the forces between between atoms increases.
neighboring atoms in a solid Fig.: (b) A graph of the 'spring' potential energy U (r)
with distance ( r) between atoms.
z Thermal Expansion is of Three Types

1. Linear Expansion
Suppose a rod of some material has a length L0 at some initial temperature T0. When the temperature
changes by T, the length changes by L. Experiments show that if T is not too large (say, less than
100C° or so), then
Change in length
Coefficient of linear expansion ( ) 
Original length × temperature difference
L

L0  T
If a body has length L0 at temperature T0, then its length L at a temperature T = T0 + T is

L = L0 + L = L0 + L0T
L = L0(1 +  T)
The unit of  is K–1 or (C°)–1.
The value of  is of the order of 10–6 to 10–5 per kelvin

2. Area Expansion or Superficial Expansion
A
Coefficient of superficial expansion () 
A0  T
If a body has area A0 at temperature T0, then its area at temperature (T0 + T)
A = A0 + A0T
A = A0 (1 + T)
3. Volume Expansion
V
Coefficient of volume expansion ( ) 
V0 T
If a body has volume V0 at temperature T0, then its volume at temperature (T0 + T) is
V = V0 (1 + T)
z Relation Between Linear Expansion, Superficial expansion and Volume Expansion

(i) For isotropic solids
 = 2
 = 3
(ii) For anisotropic solids,  = x + y + z, where x, y and z represents the mean coefficients of linear
expansion along three mutually perpendicular directions.
z Apparent Expansion of Liquids
There are two coefficients of expansion in case of liquids.

(i) Coefficient of real expansion (r)
(ii) Coefficient of apparent expansion (a)
real increase in volume
r 
Original volume × rise in temperature
apparent increase in volume
a 
Original volume × rise in temperature
with a = r – s , where s is the coefficient of cubical expansion of the solid of the vessel.
z Expansion of Cavity
If there is a hole in a plate (or cavity inside a body), the area of hole (or volume of cavity) will increase when body
expands on heating.
Two spheres, one being solid and the other hollow, are made of same material having same radius at a particular
temperature. If they are heated to the same temperature, their radius will again be same.
But if they are given the same amount of heat, the rise in temperature of the hollow sphere will be more (due
to its less mass) and hence its radius will become greater than that of the solid sphere.
z Change of Density with Temperature
Consider a body (solid) of mass m. Let V1 and 1 be its volume and density respectively at temperature
t1 and V2 and 2 are the corresponding values at temperature t2.
where T = t2 – t1
1
2 =
1  T 
 1 1  T 
–1
(as  is small)
2 = 1(1 – T)
Example 6 : A surveyor uses a steel measuring tape that is exactly 50.000 m long at a temperature of 20°C.
What is its length on a hot summer day when the temperature is 35°C? (steel = 1.2 × 10–5 K–1)
Solution : The temperature change is T = T – T0 = 15C°, so from equation (ii) the change in length L
and final length L = L0 + L are
L = L0T = (1.2 × 10–5 K–1) (50 m) (15 K) = 9.0 × 10–3 m = 9.0 mm
L = L0 + L = 50.000 m + 0.009 m = 50.009 m
Thus, the length at 35°C is 50.009 m.
This example shows that metals expand very little under moderate temperature changes. Even a
metal baking pan in a 200°C oven is only slightly larger than it is at room temperature.
Example 7 : A glass flask with volume 200 cm3 is filled to the brim with mercury at 20°C. How much mercury
overflows when the temperature of the system is raised to 100°C? The coefficient of linear
expansion of the glass is 0.40 × 10–5 K–1. Cubical expansion of mercury = 18 × 10–5 K–1.
Solution : The coefficient of volume expansion for the glass is glass = 3glass = 1.2 × 10–5 K–1
The increase in volume of the glass flask is
Vglass = glassV0T
= (1.2 × 10–5 K–1) (200 cm3) (100°C – 20°C)
= 0.19 cm3
The increase in volume of the mercury is
Vmercury = mercury V0.T
= Vmercury = mercury V0T
= (18 × 10–5 K–1) (200 cm3) (100°C – 20°C) = 2.9 cm3
The volume of mercury that overflow is
Vmercury – Vglass = 2.7 cm3
Example 8 : A pendulum clock consists of a light iron rod connected to a small, heavy bob. If it is designed
to keep correct time at 20°C, how fast or slow will it go in 24 hours at 40°C? Coefficient of linear
expansion of iron = 1.2 × 10–5/°C.
l
Solution : T = 2
g
T 1 l 1 1
        (1.2  10 5 )  (20 )
T 2 l 2 2
= 1.2 × 10–4
∵ Time period increases
 Clock goes slow
T
Hence the time lost in 24 hours = 24 hours ×
T
= (24 × 3600) × 1.2 × 10–4
= 10.4 s
EXERCISE
1. A constant volume gas thermometer shows pressure reading of 50 cm and 90 cm of mercury at 0°C and 100°C
respectively. When the pressure reading is 60 cm of mercury, the temperature is
(1) 25°C (2) 40°C
(3) 15°C (4) 12.5°C

2. On centigrade scale the temperature of a body increases by 30 degrees. The increase in temperature on
Fahrenheit scale is
(1) 50° (2) 40°
(3) 30° (4) 54°
3. A bar of iron is 10 cm at 20°C. At 19°C it will be (Fe = 11 × 10–6/°C)
(1) 11 × 10–6 cm longer (2) 11 × 10–6 cm shorter
(3) 11 × 10–5 cm shorter (4) 11 × 10–5 cm longer
4. Coefficient of volume expansion of mercury is 0.18 × 10–3/°C. If the density of mercury at 0°C is 13.6 g/cc,
then its density at 200°C is
(1) 13.11 g/cc (2) 52.11 g/cc
(3) 16.11 g/cc (4) 26.11 g/cc
5. A metre rod of silver at 0°C is heated to 100°C. It’s length is increased by 0.19 cm. Coefficient of volume
expansion of the silver rod is
(1) 5.7 × 10–5 /°C (2) 0.63 × 10–5 /°C
(3) 1.9 × 10–5 /°C (4) 16.1 × 10–5/°C
6. A solid ball of metal has a concentric spherical cavity within it. If the ball is heated, the volume of the cavity
will
(1) Increase (2) Decrease
(3) Remain same (4) All of these
7. Two rods of different materials having coefficient of thermal expansion 1 and 2 and Young’s modulus Y1 and
Y2 respectively are fixed between two rigid massive walls. The rods are heated such that these undergo same
increase in temperature. There is no bending of the rods. If 1: 2 = 2 : 3, the thermal stress developed in
the two rods are equal, provided Y1 : Y2 is
(1) 2:3 (2) 1:1
(3) 3:2 (4) 4:9
8. A glass flask of volume 200 cm3 is just filled with mercury at 20°C. The amount of mercury that will overflow
when the temperature of the system is raised to 100°C is (glass = 1.2 × 10–5/C°, mercury = 1.8 × 10–4 /C°)
(1) 2.15 cm3 (2) 2.69 cm3
(3) 2.52 cm3 (4) 2.25 cm3
9. Solids expand on heating because
(1) Kinetic energy of atom increases
(2) Potential energy of atom increases
(3) Total energy of atom increases
(4) The potential energy curve is asymmetric about the equilibrium distance between neighbouring atoms
SPECIFIC HEAT CAPACITY
The amount of heat, required to raise the temperature of unit mass, of the substance by 1°C, called specific
heat capacity or specific heat
dQ
Specific heat c 
mdT
dQ = mcdT

Q  mcdT
Value of c depends on material of body and temperature range.
cal
Specific heat of water  1 from 14.5°C to 15.5°C
gK
J
In SI system specific heat of water  4200
kg K
z Molar specific heat or molar heat capacity : Molar specific heat is defined as the amount of heat required
to raise the temperature of one mole of a substance through 1°C.
1 Q 
n  T 
Molar heat capacity =
The SI unit of C is J/mol K.

z Heat capacity or Thermal capacity of a body : Heat capacity of a body is defined as the amount of heat
required to raise the temperature of body by one degree. It is denoted by C
Heat capacity C = mc
The SI unit of heat capacity is J/K.
z Water Equivalent : Water equivalent of a body is the quantity of water whose temperature would be raised
through 1°C (or 1K) by the same amount of heat as required to raise the temperature of the body through 1°C
(or 1 K).
If w be the water equivalent of a body of mass m and specific heat capacity C, then
w × Cw × 1 = m × C × 1
mC
or, w 
Cw
In cgs system, Cw = 1 cal g–1°C–1, so w = mC
mC
In SI system, Cw = 4200 J kg–1 k–1, so w 
4200
w is measured in g in cgs units and in kg in SI units.
z Latent Heat : It is the amount of heat required to change the state of a unit mass of a substance without
change in its temperature, and pressure.
Q = mL
where L is the latent heat. Its SI unit is J/kg.
It depends on
(i) Pressure, Its value is usually quoted at standard atmospheric pressure, with increase in pressure latent
heat of vaporisation will increase, and latent heat of fusion will decrease for water.
(ii) Nature of the phase change
(iii) Properties of the substance.
The latent heat of fusion (Lf) is the heat gained when the phase change is from a solid to a liquid.
For water at 1 atm, its value is 79.7 cal/g = 3.33 × 105 J/kg. It is called latent heat of ice.
The latent heat of vaporisation (Lv) corresponds to liquid to vapour phase change. Its value for water at
1 atm is 540 cal/g = 226 × 105 J/kg. It is also called the latent heat of steam.
Note : (1) Specific heat of water is
sw  4186 J kg1K 1  1 calg1 (C)1 .

(2) Specific heat capacity of ice (–10°C) is about 2220 J/kg K.
(3) The molar heat capacities for most elemental solids are about 25 J/mol K. This is known
as the rule of Dulong and Petit (for its discoverers). This means the heat required for a
given temperature increase depends only on how many atoms the sample contains and not
on the mass of an individual atom.
(4) The specific heat at extremely low temperatures varies as the cube of the absolute
temperature. This is called Debye’s law (CT3).
(5) Hydrogen has the largest specific heat capacity among all solids, liquids and gases.
CALORIMETRY
As we know that heat is a form of energy and as such is a measurable quantity. Calorimetry deals with the
measurement of heat. The vessel which is largely used in such a measurement is called a calorimeter.
Principle of Calorimetry : When two bodies at different temperatures are allowed to share heat, they attain
a common temperature. If it is assumed that no heat is received from or given to any body from outside the
system and if there is no chemical action involved in the process of sharing, then
Heat gained by cold body = Heat lost by hot body
CHANGE OF STATE
Phase change and latent heat
Let us consider a container filled with 1 kg of ice at temperature (– 25°C).
Figure is a graph which shows the changes in the state of ice with time as more and more heat is added to it.
e
Temperature
(°C) Steam
water boils 3
100 c [3 > 1 >  2 ]
Q4 d
50 Q3
water
a Q2 b 2
O
ice
Q1
melt
1
– 25
0 t1 t2 t3 time (minute)
Fig.: Graph of temperature vs time for a specimen

of water initially in the solid phase (ice)
If sufficient amount of heat Q is added, then this heat is used up in following steps.
(i) First the temperature of whole ice will raise to 0°C i.e melting point of ice, by using heat
Q1 = mci T = 1 × 0.5 × 25 = 12.5 kcal
(ii) On supplying extra heat temperature remains at 0°C, and ice start to melt, by using heat
Q2 = mLf = 1 × 80 = 80 kcal
(iii) When whole ice turned into water, on supplying extra heat the temperature rises once again up to 100°C
(boiling point of water) by using heat
Q3 = mcwT = 1 × 1 × 100 = 100 kcal
(iv) When temperature 100°C is reached, the temperature again stays constant, and by using heat of
vaporisation whole water convert into vapours at 100°C.
Q4 = mLv = 1 × 540 = 540 kcal
(v) When all the water convert into steam, the temperature again increases.
Temperature Phase Temperature Phase
change Ice change Water change Water change Steam
lce
–25°C at at at at
Q1 = mci T 0°C Q2 = mLf 0°C Q3 = mcw T 100°C Q4 = mLv 100°C
Example 9 : 5 g ice at 0°C is mixed with 1 g steam at 100°C. Find the final temperature and composition of
the mixture.
Solution : Heat required to melt ice at 0°C = Q1 = mi Lf = 5 g × 80 cal/g = 400 cal
Heat required to raise the temperature of water from 0 to 100°C = Q2 = miswt
1 cal
= 5g × × 100°C = 500 cal.
g°C
Maximum heat ice can absorb for steam to reach boiling point is = Q1 + Q2 = 900 cal
Heat rejected by steam on complete condensation = Q3 = msLv = 1 g × 540 cal/g = 540 cal
This heat rejected can melt the ice completely but cannot raise the temperature of water from
0 to 100°C as there is a deficiency of heat = Q1 + Q2 – Q3 = (900 – 540)cal = 360 cal
Heat deficient
 Resulting temperature = 100°C –
Thermal capacity of system i.e., 6 g water
360 cal
= 100°C –
6 g  1 cal/g °C
360C
= 100°C –
6
= 100°C – 60°C = 40°C
Example 10 : 100 g ice at 0°C is mixed with 10 g steam at 100°C. Find the final temperature and composition.
Solution : Heat required to melt ice at 0°C = Q1 = miLf = 100 × 80 cal/g = 8000 cal
Heat required to raise the temperature of water from 0°C to 100°C = Q2 = miswt
= 100 g × 1 calg°C × 100°C = 10,000 cal
Maximum heat ice can absorb from steam to reach boiling point = Q1 + Q2 = 18000 cal
Heat rejected by steam on complete condensation
Q3 = msLv = 10 g × 540 cal/g = 5400 cal
Heat rejected by water at 100°C cooled to 0°C
Q4 = msswt = 10 g × 1 cal/g°C × 100 = 1000 cal

Maximum heat can be supplied to steam to ice at 0°C = Q3 + Q4 = (5400 + 1000)cal = 6400 cal
To melt the ice 8000 cal heat is required, but maximum heat supplied by steam 6400 cal is in
sufficient to melt the ice. So resulting temperature of mixture is = 0°C.
Heat supplied to steam

Amount of ice melted =
Lf
6400 cal
= = 80 g
80 cal/g
 Ice remains in the mixture = 100 g – 80 g = 20 g
Water present in the mixture = 10 g + 80 g = 90 g
Example 11 : 30 g ice at 0°C is mixed with 25 g steam at 100°C. Find the final temperature and composition.
Solution : Heat required to melt ice at 0°C = Q1 = miLf = 30 g × 80 cal/g = 2400 cal
Heat required to raise the temperature of water at 0°C to 100°C
Q2 = miswt
= 30 g × 1 cal/g°C × 100°C = 3000 cal
Maximum heat ice can absorb from steam to reach 100°C = Q3 = msLv = 25 g × 540 cal/g
= 13500 cal
Total steam cannot be condensed as ice can take a maximum of 5400 cal out of 13500 cal
 Resulting temperature of mixture = 100°C
Maximum heat absorbed by ice 5400 cal

Steam condensed =  = 10 g
Lv 540 cal/g
Steam present in the mixture = 25 g – 10 g = 15 g
Water present in the mixture = 30 g + 10 g = 40 g
Example 12 : Liquids A and B are at 30°C and 20°C. When mixed in equal masses, the temperature of the
mixture is found to be 26°C. Their specific heats are in the ratio of
(1) 3:2
(2) 1:1
(3) 2:3
(4) 4:3
Solution : Let the specific heat of A and B be SA and SB. Now heat lost by A in cooling from 30°C to 26°C
will be same as liquid B warming up from 20°C to 26°C. Since the masses are same
 m.SA (30°C – 26°C) = m.SB(26°C – 20°C)
or 4SA = 6SB
SA 6 3
or S  4  2
B
z Pressure Temperature (P-T ) Diagram and Triple Point
A graph between pressure (P) and temperature (T) of a substance is called P-T diagram. This diagram is also
called phase diagram since all the three phases are separated from each other by three lines.
(i) The boiling point of a liquid also depends upon pressure. In figure, curve OC shows how the boiling point
of water varies with pressure (behaviour typical of other liquid as well).
P (atm)
B
Critical point
218 C OA : Sublimation curve
Solid Liquid OB : Fusion curve
1
OC : Vaporisation curve
0.006 Vapour
O Triple
point
T (°C)
A 0 0.01 100 374.14
Fig.: Phase diagram for water

A curve which shows the variation of boiling point with pressure is called the vaporisation curve. In case
of water, this curve (i.e., OC) is also called steam line. The points on this curve represents the states
in which liquid and vapour phases coexist.
The upper limit (C) of this curve occurs at a temperature of 374.14°C and a pressure of 218 atm and is
known as the critical point. A substance cannot exist in the liquid state at a temperature above that of
its critical point, regardless of how great the pressure may be.
(ii) The melting point of a solid also depends upon pressure (although to a smaller extent than the boiling
point). A curve which shows the variation of melting point with pressure is called the fusion curve.
In case of water, this curve is also called the ice line. In figure, the curve OB represents the fusion curve
of water. It is to be noted that this curve leans slightly to the left. This implies that as the pressure is
increased, the melting point is lowered.
(iii) The fusion and the vaporisation curves of water intersect at a temperature of 0.01°C and at a pressure
of 0.006 atm as shown in Figure. Along the fusion curve, both ice and water can simultaneously exist.
Hence, under conditions corresponding to those of the intersection of the two curves, the solid, liquid and
gas can all exist together. This intersection is accordingly called the triple point (O) of water.
(iv) At pressure below that of its triple point, no substance can exist as a liquid. The dividing line on a
pressure-temperature graph between the solid and the vapour state is called the sublimation curve since
it represents the conditions required for a solid to vaporise directly or a vapour to solidify directly.
Triple point of a substance is that point on the pressure temperature (P-T) diagram, whose coordinates
give the particular temperature and pressure at which the solid, liquid and vapour phases of the substance
coexist in equilibrium without any change in their proportions.
Summary of Shortcuts
(i) When ice is dominant i.e. the ratio of ice at 0°C and steam at 100°C is 8 : 1 or more, entire steam
will be cooled upto 0°C water and final temperature will be 0°C. Excess of 8 : 1 ratio, ice will be
left and we’ll have ice water mixture.
(ii) When steam is dominant i.e. the steam to ice ratio is 1 : 3 or more; ice will be heated upto 100°C
water. When steam will be more than 1 : 3 ratio, excess steam will be left and we’ll have water
steam mixture at 100°C.
(iii) When the ice to steam ratio is in between 8 : 1 and 3 : 1, make conventional calculation and we’ll
have all water somewhere between 0°C to 100°C.
EXERCISE
10. 300 gm of water at 25°C is added to 100 gm of ice at 0°C. The final temperature of the mixture is
5 5
(1) – C (2) – C
3 2
(3) – 5°C (4) 0°C
11. Two spheres made of same substance have diameters in the ratio 1 : 2. Their thermal capacities are in the
ratio of
(1) 1:2 (2) 1:8
(3) 1:4 (4) 2:1
12. 80 gm of water at 30°C is poured on a large block of ice at 0°C. The mass of ice that melts is
(1) 30 gm (2) 80 gm
(3) 1600 gm (4) 150 gm
13. Work done in converting 1g of ice at –10°C into steam at 100°C is
(1) 3.04 kJ (2) 6.05 kJ
(3) 0.721 kJ (4) 0.616 kJ
14. 2 gm of steam condenses when passed through 40 gm of water initially at 25°C. The condensation of steam
raises the temperature of water to 54.3°C. What is the latent heat of steam?
(1) 540 cal/g (2) 536 cal/g
(3) 270 cal/g (4) 480 cal/g
15. The temperature of 100 gm of water is to be raised from 24°C to 90°C by adding steam to it. The mass of
the steam required for this purpose is
(1) 20 g (2) 15 g
(3) 12 g (4) 18 g
16. 10 gm of ice at –20°C is kept into a calorimeter containing 10 gm of water at 10°C. The specific heat of water
is twice that of ice. When equilibrium is reached, the calorimeter will contain
(1) 20 gm of water (2) 20 gm of ice
(3) 10 gm ice and 10 gm of water (4) 5 gm ice and 15 gm water
17. Water falls from a height 500 m. What is the rise in temperature of water at bottom if whole energy remains
in the water ?
(1) 0.96°C (2) 1.02°C
(3) 1.16°C (4) 0.23°C
18. Latent heat of ice is 80 cal/gm. A man melts 60 gm of ice by chewing in 1 minute. His power is
(1) 4800 W (2) 336 W
(3) 1.33 W (4) 0.75 W
HEAT TRANSFER
The three mechanisms of heat transfer are conduction, convection and radiation. Conduction occurs within a
body or between two bodies in contact. Convection depends on motion of mass from one region of space to
another. Radiation is heat transfer by electromagnetic radiation, such as sunshine, with no need for matter to
be present in the space between bodies.
z Conduction : The phenomena of heat transfer in which heat flows from one place to other without actual transfer
of atoms called conduction.
Most of the metals also use another, more effective mechanism to conduct heat. Within the metal, some
electrons can leave their parent atoms and wander through the crystal lattice. These “free” electrons can rapidly
carry energy from the hotter to the cooler regions of the metals, so metals are generally good conductors of heat.
Consider a rod of length L, area of cross-section A. The temperature of two ends are T1 and T2 (T1 > T2).
T1 T ( T – dT ) T2
A
x dx x= L
x=0
Consider a cross-section, at co-ordinate x its temperature is T, and at (x + dx), temperature is (T - dT)
dT
Temperature gradient = 
dx
dQ dT
In variable state rate of heat flow or heat current H   KA
dt dx
The negative sign, shows that heat always flows, in the direction of decreasing temperature
Q KA(T1  T2 )
In steady state rate of heat flow 
t L
KA(T1  T2 )t
Total heat flow in time t, Q 
L
W
Where K is coefficient of thermal conductivity, Sl unit of K is
mK
z Comparison between electrical conduction and heat conduction
Electricity Heat
(1) In metals charge carriers are free electrons. (1) In metals heat carriers are free electrons.
dq dQ
(2) Electric current i  (2) Heat current iH 
dt dt
(3) Electric potential (V) (3) Temperature (T)
(4) Charge flows due to potential difference (4) Heat flows due to temperature difference
V T
(5) Electrical resistance R  (5) Heat resistance RH 
i iH
L dQ KA(T1  T2 )
R iH  
A dt L
L T1  T2 L
R 
A iH KA
L
RH 
KA
(6) Combination of resistance (6) Combination of heat resistance

Current is same Heat current will be same
(i) Series (i) Series
Potential will divide Temperature will divide
i R1 i R2 R3 iH R1 R2 R3
V1 V2 iH
Effective resistance R = R1 + R2 + .... Effective heat resistacne RH = R1 + R2 + ....

Effective conductivity in series
d1  d 2  d3  ....
K
d1 d 2 d3
   ....
k1 k2 k3
Current will divide Heat current will divide

(ii) Parallel (ii) Parallel
Potential difference
will be same
1 1 1
Effective resistance Effective heat resistance R  R  R  ....
H 1 2
1 1 1
   .... Effective thermal conductivity in parallel
R R1 R2
K1A1  K 2 A2  ....
K eff 
A1  A2  ....
Kirchhoff’s laws and principle of wheat stone bridge also valid in heat conduction.
Wiedmann - Franz Law

The ratio of thermal and electrical conductivities (K & ) at a particular temperature (T K) is same for all metals
(except mica)
K
i.e.,  constant
T
Hence, a good electrical conductor is also a good thermal conductor. But good thermal conductors
need not be good electrical conductors e.g., sand, mica.
Example 13 : A steel bar 10.0 cm long is welded end to end to a copper bar 20.0 cm long. Both bars are
insulated perfectly on their sides. Each bar has a square cross-section, 2.00 cm on a side. The
free end of the steel bar is maintained at 100°C by placing it in contact with steam, and the free
end of the copper bar is maintained at 0°C by placing it in contact with ice. Find the temperature
at the junction of the two bars and the total rate of heat flow.
Steel T Copper
Solution : T1 = 100°C 2.00 cm T2 = 0°C
10.0 cm 20.0 cm
In the steady state, the two heat currents are equal. We have
K steel A 100C – T  K copper A T – 0C 

Hsteel = and Hcopper 
Lsteel Lcopper
The areas A are equal and may be cancelled out.
Substituting Lsteel = 0.100 m, Lcopper = 0.200 m, and numerical values of K from the previous table,
we find
 50.2 W/m.K  100C – T    385 W/m.K  T – 0C

0.100 m 0.200 m
Solving for T, we obtain T = 20.7°C
We can find the total heat current by substituting this value for T,
 50.2 W/m.K   0.0200 m2  100C – 20.7C 

Hsteel =
0.100 m
Hsteel = 15.9 W
∵ Hsteel = Hcopper
 Hcopper = 15.9 W
Example 14 : Three identical rods have been joined at a junction to make it a Y shape structure. If two free ends
are maintained at 60°C and the third end is at 0°C, then what is the junction temperature ?
60°C

0°C
60°C
Solution :
60°C
H2
H
0°C

H1
60°C
Rods are in steady state
So, H = H1 + H2
KA(  0) KA(60  ) (60  )

   KA
L L L
  = 60 –  + 60 – 
 3 = 120
  = 40°C
Example 15 : Three rods of same cross section but different length and conductivity are joined in series. If the
temperature of the two extreme ends are T1 and T2 (T1 > T2) find the rate of heat transfer H.
T1 T2
k1 k2 k3
L1 L2 L3
L1 L L
Solution : Total thermal resistance R = R1 + R2 + R3 =  2  3
k1A k2 A k3 A
T1 – T2
now, H 
R

T1 – T2 
H
L1 L L
 2  3
k1A k 2 A k3 A
EXERCISE
19. Three identical thermal conductors are connected as shown in figure. Consider no heat lost due to radiation,
the temperature of the junction is
60°C
20°C
70°C
(1) 60°C (2) 20°C
(3) 50°C (4) 10°C

20. Consider a compound slab consisting of two different materials having equal thickness and thermal
conductivities K and 2K in series. The equivalent conductivity of the slab is
2
(1) K (2) 2K
3
4
(3) 3K (4)  3 K
 
21. The outer faces of a rectangular slab made of equal thickness of iron and brass are maintained at 100°C and
0°C respectively. The temperature at the interface is
(Thermal conductivity of iron and brass are 0.2 and 0.3 respectively.)
(1) 100°C (2) 40°C
(3) 50°C (4) 70°C
22. Surface of the lake is at 2°C. The temperature of the bottom of the lake is
(1) 2°C (2) 3°C
(3) 4°C (4) 1°C

23. A body of length 1 m having cross-sectional area 0.75 m2 has heat flow through it at the rate of 6000 J/s.
The difference between two ends of conductor if K = 200 J m–1K–1 is
(1) 20°C (2) 40°C
(3) 80°C (4) 100°C
24. A slab consists of two parallel layers of two different materials of same thickness having thermal conductivities
K1 and K2. The equivalent conductivity of the combination is
K1  K2
(1) K1 + K2 (2)
2
2K1 K2 K1  K 2
(3) (4)
K1  K2 2K1K 2
25. The layers of atmosphere are heated through

(1) Convection (2) Conduction
(3) Radiation (4) Both (1) & (3)
26. The temperature gradient in a rod of 0.5 m long is 80ºC/m. If the temperature of hotter end of the rod is 30°C,
then the temperature of the colder end is
(1) 40°C (2) – 10°C
(3) 10°C (4) 0°C
27. The ratio of thermal conductivity of two rods of different material is 5 : 4. The two rods of same area of cross-
section and same thermal resistance will have the length in the ratio
(1) 4:5 (2) 9:1
(3) 1:9 (4) 5:4
28. Five rods of same dimensions are arranged as shown in the figure. They have thermal conductivities K1, K2, K3,
K4 and K5. When points A and C are maintained at different temperature, no heat flows through the central rod if
B
K4 K3
A C
K5
K2 K1
D
(1) K1 = K4 and K2 = K3 (2) K1K4 = K2K3
K1 K2
(3) K1K2 = K3K4 (4) 
K4 K3
Convection
Convection is the transfer of heat by mass motion of a fluid from one region of space to another. There are
two types of convection.
(i) Forced convection : If the fluid is circulated by an external agent like a blower or pump, the convection
process is known as forced convection. The most important mechanism for heat transfer within human
body is forced convection of blood, with the heart serving as the pump.
(ii) Natural convection or free convection : If the flow is caused by difference in density due to thermal
expansion, such as hot air rising, the process is called natural convection or free convection. Free
convection in the atmosphere plays an important role in determining the daily weather, and convection
in the oceans is an important global heat transfer mechanism.
In convection heat transfer is very complex process, and there is no simple equation to describe it. Here are
a few experimental facts:
(i) The heat current due to convection is directly proportional to the surface area. This is the reason for the
large surface areas of radiators and cooling fans.
(ii) The viscosity of fluids slows natural convection near a stationary surface, giving a surface film that on a
vertical surface typically has about the same insulating value as 1.3 cm of plywood (R value = 0.7). Forced
convection decreases the thickness of this film, increasing the rate of heat transfer. This is the reason
for the “wind chill factor”, you get cold faster in a cold wind than in still air with the same temperature.
5
(iii) The heat current due to convection is found to be approximately proportional to the power of the
4
temperature difference between the surface and the main body of fluid.
z Radiation
Radiation is the transfer of heat by electromagnetic waves such as visible light, infrared, and ultraviolet rays.
This heat transfer would occur even if there were nothing but vacuum between you and the source of heat.
Everybody, even at ordinary temperatures, emits energy in the form of electromagnetic radiation. At ordinary
temperature, say 20°C, nearly all the energy is carried by infrared radiation with wavelengths much longer than
those of visible light. As the temperature rises, the wavelength shift to shorter values. At 3000°C, the temperature
of an incandescent lamp filament, the radiation contain enough visible light that the body appears “white-hot”.
Heat radiations have all the properties of visible light except frequency and wavelength. Heat radiations follows
all the laws of visible light like, reflection, refraction, interference etc.
z Good absorbers are bad reflectors

Let Q be the radiant energy incident on the surface AB as shown in Figure
Q Q1
Q  incident energy
Q1  reflected energy
A Q2 B Q2  absorbed energy
Q3  transmitted energy
Q3
If out of this Q incident energy, Q1 is reflected, Q2 is absorbed and Q3 is transmitted, then

Q1
r (reflectance of the surface)
Q
Q2
a (absorptance of the surface)
Q
Q3
t (transmittance of the surface)
Q
None of these quantities (r, a and t) has a unit as these are pure ratios.
Thus, r + a + t = 1
In case, the surface does not transmit radiation, t = 0 and as r + a = 1
obviously, if a is more, r is less or vice-versa. Thus, good absorbers are bad reflectors and bad absorbers are
good reflectors. But good absorbers are good emitter and vice versa.
z Blackbody
A perfect blackbody is one which absorbs radiation of all wavelengths incident on it. Its absorptance is unity
as it neither reflects nor transmits any radiations.
A blackbody does not mean that its colour is black, although a black-coloured body may come close to being
a blackbody. Examples of black bodies, Fery’s black body, the sun.
z Fery’s blackbody
The radiation inside an enclosure whose inner walls are maintained at a constant temperature has the same
properties as the black body radiation.
Hole behaves like

black body
Conical
projection
Lamp black
coating
Most of the energy entering through hole is absorbed (  99%).
z Stefan’s Law
From the experimental study of rate of emission of radiations by a hot body, Stefan in 1879, proposed a law
which after him is called Stefan’s law. Boltzmann gave a theoretical proof of this law by applying the laws
of thermodynamics to the blackbody radiations. This law is, therefore, also called Stefan-Boltzmann Law.
Now, as per Stefan’s Law the radiant energy emitted by a perfectly black body per unit area per second is
directly proportional to the fourth power of its absolute temperature.
E  T4
 E = T 4
 watt 
Where E = heat radiation energy per unit time per unit area  2  = emissive power
m 
T = absolute temperature [K]
 = Stefan-Boltzmann’s constant and  = 5.67×10–8 Wm–2K–4
Rate of emission from the body
H = EA = AT 4 [A = outer surface area]
Remember : Whether the radiating body is hollow or solid it is the outer surface area only which comes
into the formula H  AT 4
For non-black bodies : The heat radiated happens to be less and for these bodies we write the equation
H  eAT 4 where, e is known as ‘emissivity’ or ‘relative emittance’ and has value 0 < e < 1 depending on
the nature of the surface.
So, emissivity e for a body is defined as
Emissive power of a body at absolute temperatureT
e=
Emissive power of black body at the same absolute temperature T
For blackbody the e = 1 Any
Body Surroundings
Let the surroundings be at temperature T0 and let a body be at temperature T0
temperature T
at temperature. T (T > T 0 ). The body emits, as well as
receives energy from surrounding. Then the net heat exchange
between the body and the surrounding per unit time will be
H = eA T – T0  .
4 4
NEWTON’S LAW OF COOLING

The rate of loss of heat of a body is directly proportional to the excess of temperature (T – T0) of the body
with respect to surroundings.
Rate of loss of heat.
dQ
  (T  T0 ) (T0 = Temperature of surroundings)
dt
dQ
  k (T  T0 ) (i) Where ( k  4T03 Ae )
dt
dT
– mc  k (T  T0 )
dt
dT k
Rate of cooling,   T  T0  (ii)
dt mc
dT k
 dt
T  T0 mc
dT k
T T 0

mcdt
k
ln(T  T0 )  Kt  c (iii) Where K  and c = constant of integration
mc
(i) For Newton’s law of cooling graph between ln (T - T0) and time t is a straight line
ln(T – T 0)
(t )
time
(ii) Newton’s law of cooling is a special case of Stefan-Boltzmann’s law applicable for small temperature
difference.
(iii) Newton’s law of cooling can also be used in form
 dT  Ti  Tf
   K Tav  T0 , Where Tav 
 dt  2
Example 16 : A body cools in 7 min from 60°C to 40°C. What will be its temperature after the next 7 minutes?
The temperature of the surroundings is 10°C.
Solution : In the First case,
 T – T0 
T1 = 60°C, T2 = 40°C, T0 = 10°C, t = 7 min from ln  1  Kt ,
 T2 – T0 
we get,
 60 – 10 
ln    7K
 40 – 10 
5
ln = 7K ...(i)
3
In the Second case,
if T is the temperature after next 7 min, T1 = 40°C, T2 = T, T0 = 10°C, t = 7 min
ln
 40 – 10
T – 10 = 7K
30
ln = 7K ...(ii)
T – 10
From equations (i) & (ii),
5 30
ln = ln
3 T – 10
5 30

3 T – 10
T = 28°C
BLACK BODY SPECTRUM

If a black body, heated to different temperatures and a graph is drawn between wavelength and intensity corresponding
to those wavelengths, then it is of following shapes. From graphs following conclusion can be drawn.
e
T3 > T2 > T 1
T3
T2
T1
m m m

3 2 1
(i) At a certain temperature as wavelength increases, intensity corresponding to those wavelength also increase,
achieves a maximum value and again start to decrease. It means at a given temperature, spectral emissive
power is maximum for a particular wavelength. Spectrum of black body is continuous spectrum.
(ii) As temperature increases, wavelength corresponding to maximum intensity, shift towards lower wavelength.
Wavelength corresponding to maximum intensity is inversely proportional to absolute temperature.
1
m 
T
 mT  b (This is Wein’s displacement law)
b = Wein’s constant = 2.89 × 10–3 m K.

(iii) As temperature increases, area under the graph also increases, the area under the graph, gives the total
radiated average power per unit surface area and is found proportional to forth power of absolute temperature.
Hence Stefan's Law proved.
(iv) The maximum intensity corresponding to any wavelength is proportional to fifth power of absolute temperature
(e )m  T 5 This is Wein’s energy law
z Solar Constant and the Temperature of the Sun

The amount of radiant energy received by the Earth per unit area per unit time from the Sun in the absence
of atmosphere when placed at right angles to the incident radiation at a distance equal to the mean distance
of the Earth from the Sun, is called solar constant (S). Its measured value is approximately 1360 Wm–2.
The radiant energy emitted per second by the Sun is
H = (4R2)T 4
where R is the radius of the Sun.
If the mean distance of the Earth from the Sun is r, then the radiant power received per unit area is
2
H 4R 2T 4  R 
Solar constant (S) = 2 =    T 4 R = Radius of sun.
4r 4r 2 r
R
Here = mean angle subtended by the solar radius at the Earth = 4.65 × 10–3 rad.
r
2
R
Temperature of the Sun : From S    T 4 ,
r 
1/4
 r 2S 
T  2 
 R  
EXERCISE
29. In heat transfer, which method is based on gravitation
(1) Natural convection (2) Conduction
(3) Radiation (4) All of these

30. In which process, the rate of transfer of heat is maximum?
(1) Conduction (2) Convection
(3) Radiation (4) In all these, heat is transferred with the same velocity
31. Good absorbers of heat are
(1) Poor emitters (2) Non-emitters
(3) Good emitters (4) Highly polished

32. A body, which emits radiations of all possible wavelengths, is known as
(1) Good conductor (2) Partial radiator
(3) Absorber of photons (4) Perfectly black body
33. A hot and a cold body are kept in vacuum separated from each other. Which of the following causes decrease
in temperature of the hot body?
(1) Radiation (2) Convection
(3) Conduction (4) Temperature remains same
34. A liquid cools down from 70°C to 60°C in 5 minutes. The time taken to cool it from 60°C to 50°C will be
(1) 5 minutes
(2) Lesser than 5 minutes
(3) Greater than 5 minutes
(4) Lesser or greater than 5 minutes depending upon the density of the liquid
35. If a metallic sphere gets cooled from 62°C to 50°C in 10 minutes and in the next 10 minutes gets cooled to
42°C, then the temperature of the surroundings is
(1) 30°C (2) 36°C
(3) 26°C (4) 20°C

36. Newton’s law of cooling is used in laboratory for the determination of the
(1) Specific heat of the gases (2) The latent heat of gases
(3) Specific heat of liquids (4) Latent heat of liquids

37. It takes 10 minutes to cool a liquid from 61°C to 59°C. If room temperature is 30°C then time taken in cooling
from 51°C to 49°C is
(1) 10 min (2) 11 min
(3) 13 min (4) 15 min
38. A block of metal is heated to a temperature much higher than the room temperature and allowed to cool in a
room free from air currents. Which of the following curves correctly represents the cooling? (T : Temperature
of block)
T T
(1) (2)
time time
T T
(3) (4)
time time
SOME IMPORTANT FORMULAE
1. Bimetallic Strip
A bimetallic strip consists of two strips of equal length but of different metals, riveted together keeping one
over the other.
Fe
Fe
Cu
Cu
1
When such a bimetallic strip is heated, it bends with metal of greater  on outer side, i.e., convex side.
This bimetallic strip is specially used in thermostat or auto-cut in electric heating circuits. It has also been
used as thermometer by calibrating its bending.
2. Expansion in Metallic Scales

If a scale say of steel gives correct reading at temperature 0 then at temperature  (> 0) due to linear
expansion of scale reading will be lesser than true value. Also, if the measured object (say copper) also
expands on heating then
1  s   – 0  
True value = Scale reading
1  Cu   – 0  
3. Effect of Linear Expansion on Pendulum Clocks

Suppose a pendulum clock gives proper time at temperature  and length of its pendulum at this temperature
is L. If temperature is increased to , then due to linear expansion, length of pendulum and hence its time
period will increase. Let L ang T be its length and time period at temperature .
Let t be change in time of clock in time interval t
1
So, t =  t
2
1
(a) Hence, time lost by the clock in a day (86,400 s) =   86,400   43200 s
2
(b) Clock will gain time, i.e., clock will become fast if  < .
(c) The gain or loss in time is independent of time period T and depends on the time interval t.
4. Formation of Ice on The Surface of Ponds or Lakes. Air (–°C)
– °C = temperature of the atmosphere
x ice
 = density of ice
L = Latent heat of fusion of ice
K = Thermal conductivity of ice dx
L
or, t ( x22  x12 ) x1 = Initial thickness water (0°C)
2K 
x2 = Final thickness
5. Effect on Buoyancy
The thrust on V volume of a body in a liquid of density  is given by B = Vg. Now with rise in temperature by
 °C, due to expansion, volume of body will increase while density of liquid will decrease according to the
relations.
V  = V(1 + s)

and  
1    
B V g 1   s  
    1    s –   
B V g 1    
(Using binomial approximation)
Now as s < , B  < B i.e., with rise in temperature thrust also decreases. This in turn implies that if a body is
weighed in a liquid at different temperatures, with rise in temperature due to decrease in thrust its weight
(= W0 – B) will increase.

t
en
nm nment
sig ssig
As A Assignment
Assignment
6. 30 g ice at 0°C is mixed with 2 g steam at 100°C.
SECTION - A The final temperature of the mixture is
NCERT Based MCQs [NCERT Pg. 289]
1. Steam is passed through 40 g of ice at (1) 0°C
–10°C till temperature of mixture becomes 80°C.
(2) 100°C
Mass of mixture will be approximately
(3) 50°C
[NCERT Pg. 289]
(4) 65°C
(1) 12 g (2) 52 g
7. Three identical conductors, P, Q and R are
(3) 45 g (4) 55 g symmetrically fixed at point O as shown in figure.
2. The temperature at which Fahrenheit reading is The temperature of junction O is
five times of Celsius reading is [NCERT Pg. 279] [NCERT Pg. 292]
(1) 10°C
(2) 12°C 60°C
40°C
(3) 18°C P
Q
(4) –10°C
O
3. It takes 10 minutes to cool a liquid from 65° to
R
55°C. If room temperature is 30°C then time taken
in cooling from 51° to 49°C is [NCERT Pg. 296]
50°C
(1) 2 minutes
(1) 60°C (2) 40°C
(2) 3 minutes
(3) 50°C (4) 45°C
(3) 1 minute
8. The thermal capacity of 50 g of aluminium is
(4) 5 minutes
(Specific heat = 0.2 cal/g°C) [NCERT Pg. 284]
4. The Wein’s displacement law express, the relation (1) 20 cal/°C (2) 200 cal/°C
between [NCERT Pg. 294]
(3) 10 cal/°C (4) 100 cal/°C
(1) Radiation energy and wavelength
9. Which of the following instruments can be used to
(2) Temperature and speed of light measure the temperature of sun?
(3) Colour of light and wavelength [NCERT Pg. 279]
(4) Wavelength corresponding to maximum (1) Pyro-heliometer
intensity and temperature
(2) Magnetic thermometers
5. A black body which is at a high temperature T K, (3) Resistance thermometers
emitted thermal radiation at the rate of E watt/m2.
The thermal radiation emitted by a body of (4) Thermoelectric thermometers
emissivity equal to 0.5 at temperature T/4 (in 10. When water is heated from 0°C to 10°C, then its
watt/m2) is [NCERT Pg. 295] density [NCERT Pg. 282]
(1) E/512 (1) Increases
(2) E/4 (2) Decreases
(3) E/256 (3) First increases then decreases
(4) E/128 (4) First decreases then increases
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456
11. The maximum wavelength of black body emission

SECTION - B
m, changes with absolute temperature T of body as
[NCERT Pg. 294] Objective Type Questions
m m 1. A faulty thermometer reads melting point of water

as 10° and boiling point as 190º. How much will
the faulty thermometer read when the actual
(1) (2) temperature is 60ºC?
T T (1) 108º
m m (2) 118º
(3) 128° (4) 138º
2. It is given that the specific heat of copper is one
(3) (4)
tenth that of water. A copper ball of 100 g at 200ºC
T T is put into 100 cc water at 24°C. Assuming no
12. In natural convection, a heated portion of liquid heat loss to the surroundings. The final
moves because of [NCERT Pg. 293] temperature of the system is
(1) Density difference from surrounding (1) 60°C
(2) Molecular collision within it (2) 55°C
(3) Small conduction loop forms (3) 40°C
(4) Its molecular motion becomes aligned (4) 35ºC

13. Two cylinders of same diameter and length one of 3. A bimetallic strip is two different metal bars 1 and
iron and other of silver are placed in close contact 2 welded parallel to their length. Given thermal
as shown. If thermal conductivity of silver is 11 coefficient of linear expansion 1 > 2 and length of
times than that of iron, then temperature of bars is same. Upon cooling the strip will bend with
interface A is approximately [NCERT Pg. 292]
(1) Concave on side 2
Fe A Ag (2) Concave on side 1
100°C 0°C (3) Convex on side 1
(1) 91.7° (2) 80°C (4) Any of the above is possible
(3) 8.3°C (4) 50°C 4. A solid metal sphere is rotating about its diameter
with constant angular speed . If its temperature is
14. The thermal conductivity of a rod depends on
increased appreciably, then its rotational speed will
[NCERT Pg. 292]
(1) Increase
(1) Material of rod
(2) Decrease
(2) Length
(3) Remain constant
(3) Mass
(4) Increase then will become constant
(4) Cross sectional area of rod
5. The thermal coefficient of linear expansion of an
15. A body cools in 4 minute from 60°C to 50°C. The anisotropic solid metal along x, y, z directions are
temperature of the body after the next 4 minute is x = 2 × 10–5 per ºC y = 3 × 10–5 per ºC and
(The temperature of the surroundings is 10°C) z = 4 × 10 –5 per ºC respectively. Its thermal
[NCERT Pg. 296] coefficient of volume expansion  should be
(1) Greater than 40°C but less than 50°C (1) 6 × 10–5 per ºC
(2) 40°C (2) 7 × 10–5 per ºC
(3) Less than 40°C (3) 8 × 10–5 per ºC
(4) Less than 40°C but greater than 35°C (4) 9 × 10–5 per ºC
6. A steel measuring tape and a copper rod are at 10. A pressure cooker having ice at – 20ºC is heated
their true length at room temperature. Their thermal at constant heat supply rate till the steam comes
coefficient of linear expansion are s and c out at 350ºC. The possible graphical variation of
respectively. However measurement of copper rod temperature T of cooker with time t is best shown
length by steel tape is shown as L 0 at a by (Assume ideal situation)
temperature  higher than room temperature. The
correct length of the copper rod at room T T
temperature is
(1) L0(1 + s .) (1) (2)
(2) L0(1 + c .) t t
(3) L0[1 + (s + c )] T T
(4) L0[1 + (s – c )]

(3) (4)
7. Inside a solid metal cube there is a spherical
t t
cavity. If the cube is heated, then volume of
(1) Cube as well as cavity decreases 11. Three rods of identical geometry but different
materials are welded to form the english alphabet
(2) Cube as well as cavity increases
letter Y as shown. If their conductivity are K, 2K and
(3) Cube increases but of cavity decreases 3K and end temperature are 20ºC. 20ºC and 100ºC
(4) Cube decreases but of cavity increases respectively, then their junction temperature Tj is
8. A cylindrical metal rod of length L, Young’s 20ºC
modulus of elasticity Y and thermal coefficient of K
linear expansion  is fixed between two concrete Tj
walls. The cross-sectional area of the rod is A. If
the temperature of the rod is increased by , then
100ºC 3K
the compressive force developed in the rod can be 2K
written as
20ºC
YA  (1) 70ºC (2) 60ºC
(1)
1   (3) 50ºC (4) 40ºC
(2) YA   
2 12. For water at 0ºC it takes one hour for first 3 mm
of ice layer formation. How much time will it take
YA 2 for next 6 mm of ice formation?
(3)
1   (1) 2 hour (2) 4 hour
YA (3) 6 hour (4) 8 hour
(4)
1  2 13. A solid sphere, solid cylinder and circular disc of
9. The apparent volume expansion coefficient of a same material and mass are maintained at same
liquid in steel vessel is A and that in aluminium temperature. The body which is cooling at fastest
vessel is B. If the thermal coefficient of linear rate is
expansion of aluminium is C, then that of steel will (1) Solid sphere (2) Solid cylinder
be
(3) Circular disc (4) All same
A  B  3C 14. Choose the correct statement
(1)
3 (1) Gas thermometers can be used for calibrating
B  A  3C other thermometers
(2)
3 (2) Platinum resistance thermometers are based
A  B  3C on Wheatstone bridge
(3) (3) Optical radiation pyrometer are based on
3
Stefan’s law
B  A  3C
(4) (4) All of these
3
15. Find the temperature at which Fahrenheit reading 20. A body at temperature T is getting cooled in an
is 1.5 times the Celsius reading. atmosphere of temperature T0. Assuming Newton’s
law of cooling to be valid, which of the following
(1) – 107º F
graphs shows the vartiation of ln(T-T0) with time t?
(2) – 160º F
ln (T-T0)
ln (T-T0)
(3) – 160º C
(4) 160º F
(1) (2)
16. Which of the following graphs correctly represents
the relation between Celsius (ºC) and Fahrenheit t t
(ºF) scale?
ln (T-T0)
ln (T-T0)
°C °C
(3) (4)
(1) (2) t t
21. In a constant volume gas thermometer the pressure
°F °F of the gas in the bulb is 28 cm and 44 cm of Hg
at temperature 0°C and 100°C respectively.
°C °C Pressure of the bulb at temperature 50°C will be
(1) 32 cm of Hg
(2) 34 cm of Hg
(3) (4)
(3) 36 cm of Hg
°F °F (4) 38 cm of Hg
17. 100 gram ice at 0°C is mixed with 20 gram steam 22. Two bodies X and Y of identical geometry, mass
at 100°C. Assuming no heat loss to the and surface finish but different material are getting
surrounding, the final temperature of the mixture is cooled in an atmosphere. Variation of their
temperature T with time t is as shown. If SX and
(1) 30°C (2) 40°C SY are their specific heat then
(3) 50°C (4) 60°C T
18. Two conducting rods are joined in series as shown y
in the diagram. Their conductivity are K and 2K.
x
Their length are in the ratio 2L and L respectively.
t
If extreme ends are maintained at 100°C and 0°C,
(1) SX = SY
then the junction temperature Tj is (Assume one
dimensional steady state heat transfer) (2) SX > SY
(3) SX < SY
100°C Tj 0°C 1
(4) S X  S
Q K 2K Q Y
2L L 23. Assume that in case of black body radiation the

product of absolute temperature T of body and the
(1) 80°C (2) 60°C
wavelength  for maximum spectral intensity of
(3) 40°C (4) 20°C radiation is constant, then which of the following
19. A cup of tea cools from 80ºC to 79.9°C in 5 minute. colors of black bodies will have highest
If temperature of surroundings is 20°C. Then how temperature?
much time will it take to cool from 70°C to 69.9°C? (1) Blue
Assume Newton’s law of cooling is valid here (2) Green
(1) 5 minute (2) 6 minute (3) Yellow
(3) 7 minute (4) 8 minute (4) Red
24. M kilogram of ice is being heated from –T1°C to (1) 25°C (2) 20°C
T2°C and variation of its temperature T along with (3) 35°C (4) 40°C
heat inputs Q is shown. Latent heat of melting
process can be calculated as 28. Two coaxial cylinders of radius r and 3r are made
of material of thermal conductivity K 1 and K 2
T (Temperature)
respectively. The equivalent thermal conductivity of
T2 substance for axial heat flow may be
K2 3r
r
O Q (Heat input) K1
Q1 Q2 Q3
–T1 4(K1  K 2 ) K1  6K 2
(1) (2)
9 9
Q1 Q2  Q1 K1  8K 2 K1  K 2
(1) (2) (3) (4)
M M 9 2
Q2 Q3  Q2 29. Two metal spheres of same material have diameter
(3) (4) in the ratio of 1 : 2. The ratio of their rates of
M M
cooling is
25. Shown below is a composite rod of metal A and B.
(1) 4 : 1 (2) 2 : 1
Their thermal coefficients of linear expansion are 1
and 2. Their lengths are L1 and L2 respectively. If (3) 1 : 2 (4) 1 : 4
on heating the length of the entire composite rod
30. A body is allowed to cool. It takes 15 minutes to
does not change, then which of the following
cool from 80°C to 60°C. The temperature of body
relations is true?
in next 15 minutes would be (temperature of
A B surrounding is 20°C)
1 2
(1) 41.6°C (2) 46.7°C
L1 L2 (3) 49.2°C (4) 52°C
(1) 1 = 2 31. A small cavity of diameter 3 cm is made in a
(2) 1 L1 = 2 L2 metal piece. On heating the metal piece, the
(3) 1 L1 + 2 L2 = 0 diameter of the cavity would
(4) (1 L1)2 + (2 L2)2 = 0 (1) Increase
26. 1.2 kg ice at 0°C is mixed with 1 kg water at 24°C (2) Decrease
in a closed calorimeter. The fraction of ice that (3) Remains unchanged
melts is (Lf = 80 cal/g)
(4) First decreases then increases
2 1
(1) (2) 32. 200 cal/minute heat energy is supplied to a solid
3 3 of 100 g. The temperature of solid Q varies with
1 1 time as shown. The latent heat of fusion of the
(3) (4)
4 2 substance is
27. Four identical rods are arranged as shown in the Q
figure. The temperature of the junction O in steady (0°C)
state is
0°C Q2
Q1
10°C 30°C t
(0, 0) 5 20 30 40
O (Minute)
(1) 6 cal/g (2) 18 cal/g
(3) 24 cal/g (4) 30 cal/g
40°C
33. A pendulum clock (made of metal) gives correct 37. A black body radiation has maximum wavelength
time at 30°C. If the room temperature rises to  m at temperature 3000 K. Its corresponding
40°C then error in the clock per 10 hours would be wavelength at temperature 4000 K will be
( = 10–3/°C)
4 3
(1) 40 s (2) 120 s (1)  (2) 
3 m 4 m
(3) 180 s (4) 240 s
5 3
(3)  (4) 
34. A black body at 327°C suspended in a black 6 m 5 m
enclosure at 27°C cools at certain rate. If
38. A cylindrical rod is used to conduct heat energy.
temperature of black body is lowered to t   °C so
If it conducts Q cal/sec, then what energy will it
that rate of cooling becomes half of initial value
conduct if its all linear dimensions are halved?
then find approximate value of t in °C.
(1) 129° C (2) 239° C Q
(1) Q = 2Q (2) Q  
2
(3) 429° C (4) 512° C
Q
35. A body cools from 70ºC to 60ºC in 5 minute, then (3) Q = 4Q (4) Q  
it will cool from 60ºC to 50ºC in time (assume 4
same surrounding whose temperature is 30ºC) 39. Select correct statement regarding “Newton’s law
of cooling”
(1) 6 min. (2) 7 min.
(3) 8 min. (4) 9 min. (1) It is used to determine latent heat of solid
36. Which of the following curves, showing black body (2) It is used to determine latent heat of liquid
radiation at temperatures T1 and T2 (T2 > T1) is (3) It is used to determine specific heat of liquid
correct?
I (4) It is used to determine specific heat of gases
40. 3.2 kg of ice at –10ºC just melts with m mass of
steam at 100ºC, then m is equal to
T2  cal 
(1)  Sice  0.5 gºC 
 
T1

(1) 800 g (2) 625 g
I
(3) 425 g (4) 325 g
SECTION - C
T2
(2) Previous Years Questions
T1
 1. A deep rectangular pond of surface area A,
containing water (density = ), specific heat
I
capacity = s), is located in a region where the
outside air temperature is at a steady value of –
26°C. The thickness of the frozen ice layer in this
T2 pond, at a certain instant is x. [NEET-2019 (Odisha)]
(3)
T1 Taking the thermal conductivity of ice as K, and its
 specific latent heat of fusion as L, the rate of
I increase of the thickness of ice layer, at this
instant, would be given by
(1) 26K/x(L + 4s)
T1 (2) 26K/x(L – 4s)
(4)
T2 (3) 26K/(x2L)
 (4) 26K/(xL)
2. An object kept in a large room having air K1  K 2 3  K1  K 2 
temperature of 25°C takes 12 minutes to cool from (1) (2)
2 2
80°C to 70°C. The time taken to cool for the same (3) K1 + K2 (4) 2(K1 + K2)
object from 70°C to 60°C would be nearly,
8. Two identical bodies are made of a material for
[NEET-2019 (Odisha)] which the heat capacity increases with
(1) 15 min (2) 10 min temperature. One of these is at 100°C, while the
other one is at 0°C. If the two bodies are brought
(3) 12 min (4) 20 min
into contact, then assuming no heat loss, the final
3. A copper rod of 88 cm and an aluminium rod of common temperature is [NEET (Phase-2)-2016]
unknown length have their increase
(1) 50°C
in length independent of increase in temperature.
The length of aluminium rod is : (Cu = 1.7 × 10–5 (2) More than 50°C
K–1 and Al = 2.2 × 10–5 K–1) [NEET-2019]
(3) Less than 50°C but greater than 0°C
(1) 6.8 cm (2) 113.9 cm
(4) 0°C
(3) 88 cm (4) 68 cm
9. A body cools from a temperature 3T to 2T in
4. The unit of thermal conductivity is : 10 minutes. The room temperature is T. Assume
that Newton's law of cooling is applicable. The
[NEET-2019]
temperature of the body at the end of next
(1) J m K–1 (2) J m–1 K–1 10 minutes will be [NEET (Phase-2)-2016]
(3) W m K–1 (4) W m–1 K–1
7 3
5. The power radiated by a black body is P and it (1) T (2) T
4 2
radiates maximum energy at wavelength, 0. If the
temperature of the black body is now changed so 4
that it radiates maximum energy at wavelength (3) T (4) T
3
3
 0 , the power radiated by it becomes nP. The 10. Coefficient of linear expansion of brass and steel
4
value of n is [NEET-2018] rods are 1 and 2. Lengths of brass and steel
rods are l 1 and l 2 respectively. If (l 2 – l 1 ) is
3 4 maintained same at all temperatures, which one of
(1) (2)
4 3 the following relations holds good? [NEET-2016]
81 256 (1) 1l1 =2l2
(3) (4)
256 81
(2) 1l2 =2l1
6. A spherical black body with a radius of 12 cm
radiates 450 watt power at 500 K. If the radius (3) 1l 22   2 l12
were halved and the temperature doubled, the
power radiated in watt would be [NEET-2017] (4) 12 l 2  22 l1
(1) 225 (2) 450 11. A piece of ice falls from a height h so that it melts
completely. Only one-quarter of the heat produced
(3) 1000 (4) 1800 is absorbed by the ice and all energy of ice gets
7. Two rods A and B of different materials are welded converted into heat during its fall. The value of h is
together as shown in figure. Their thermal [Latent heat of ice is 3.4 × 105 J/kg and g = 10 N/kg]
conductivities are K 1 and K 2 . The thermal
[NEET-2016]
conductivity of the composite rod will be
(1) 68 km
[NEET-2017]
(2) 34 km
A K1
T1 T2 (3) 544 km
B K2
(4) 136 km
d
12. A black body is at a temperature of 5760 K. The 18. A piece of iron is heated in a flame. It first becomes
energy of radiation emitted by the body at dull red then becomes reddish yellow and finally turns
wavelength 250 nm is U1, at wavelength 500 nm is to white hot. The correct explanation for the above
U2 and that at 1000 nm is U3. Wien's constant, observation is possible by using [NEET-2013]
b = 2.88 × 106 nmK. Which of the following is
(1) Wien's displacement law
correct? [NEET-2016]
(1) U2 > U1 (2) U1 = 0 (2) Kirchoff's law
(3) U3 = 0 (4) U1 > U2 (3) Newton's law of cooling

13. The value of coefficient of volume expansion of (4) Stefan's law
glycerin is 5 × 10–4 K–1. The fractional change in 19. Liquid oxygen at 50 K is heated to 300 K at constant
the density of glycerin for a rise of 40°C in its pressure of 1 atm. The rate of heating is constant.
temperature is [Re-AIPMT-2015] Which one of the following graphs represents the
(1) 0.010 (2) 0.015 variation of temperature with time?
(3) 0.020 (4) 0.025 [AIPMT (Prelims)-2012]
14. The two ends of a metal rod are maintained at
Temperature
Temperature
temperatures 100°C and 110°C. The rate of heat
flow in the rod is found to be 4.0 J/s. If the ends
are maintained at temperatures 200°C and 210°C, (1) (2)
the rate of heat flow will be [AIPMT-2015]
Time Time
(1) 4.0 J/s (2) 44.0 J/s
Temperature
Temperature
(3) 16.8 J/s (4) 8.0 J/s
15. On observing light from three different stars P, Q
and R , it was found that intensity of violet colour (3) (4)
is maximum in the spectrum of P, the intensity of
Time Time
green colour is maximum in the spectrum of R and
the intensity of red colour is maximum in the 20. If the radius of a star is R and it acts as a black
spectrum of Q. If TP, TQ and TR are the respective body, what would be the temperature of the star, in
absolute temperatures of P, Q and R then it can which the rate of energy production is Q?
be concluded from the above observations that [AIPMT (Prelims)-2012]
[AIPMT-2015] 1/ 4 1/ 4
 4R 2Q   Q 
(1) TP < TQ < TR (2) TP > TQ > TR (1)   (2)  2 
    4R  
(3) TP > TR > TQ (4) TP < TR < TQ
1/ 2
Q  Q 
16. Steam at 100°C is passed into 20 g of water at (3) (4)  2 
4R 2   4R  
10°C. When water acquires a temperature of 80°C,
the mass of water present will be: ( stands for Stefan's constant)
[Take specific heat of water = 1 cal g–1°C–1 and 21. A slab of stone of area 0.36 m2 and thickness
latent heat of steam = 540 cal g–1] [AIPMT-2014] 0.1 m is exposed on the lower surface to steam at
100°C. A block of ice at 0°C rests on the upper
(1) 24 g (2) 31.5 g
surface of the slab. In one hour 4.8 kg of ice is
(3) 42.5 g (4) 22.5 g melted. The thermal conductivity of slab is (Given,
17. Certain quantity of water cools from 70°C to 60°C latent heat of fusion of ice = 3.36 × 105 J kg–1)
in the first 5 minutes and to 54°C in the next [AIPMT (Mains)-2012]
5 minutes. The temperature of the surroundings is (1) 1.24 J/(m-s-°C)
[AIPMT-2014] (2) 1.29 J/(m-s-°C)
(1) 45°C (2) 20°C (3) 2.05 J/(m-s-°C)
(3) 42°C (4) 10°C (4) 1.02 J/(m-s-°C)
22. When 1 kg of ice at 0°C melts to water at 0°C, the 27. On a new scale of temperature (which is linear) and
resulting change in its entropy, taking latent heat called the W scale, the freezing and boiling points
of ice to be 80 cal/°C, is [AIPMT (Prelims)-2011] of water are 39°W and 239°W respectively. What
will be the temperature on the new scale,
(1) 293 cal/K (2) 273 cal/K
corresponding to a temperature of 39°C on the
(3) 8 × 104 cal/K (4) 80 cal/K Celsius scale? [AIPMT (Prelims)-2008]
23. A cylindrical metallic rod in thermal contact with two (1) 139°W (2) 78°W
reservoirs of heat at its two ends conducts an
(3) 117°W (4) 200°W
amount of heat Q in time t. The metallic rod is
melted and the material is formed into a rod of half 28. Assuming the sun to have a spherical outer surface
the radius of the original rod. What is the amount of radius r, radiating like a black body at temperature
of heat conducted by the new rod, when placed in t°C, the power received by a unit surface, (normal
thermal contact with the two reservoirs in time t? to the incident rays) at a distance R from the centre
[AIPMT (Prelims)-2010] of the sun is (where  is the Stefan’s constant)
[AIPMT (Prelims)-2007]
Q Q
(1) (2)
4 16 r 2 (t  273 ) 4 4r 2 t 4
(1) (2)
Q R2 R2
(3) 2Q (4)
2
r 2 (t  273 ) 4 16  2 r 2 t 4
24. The total radiant energy per unit area, normal to the (3) 2 (4)
4R R2
direction of incidence, received at a distance R from
the centre of a star of radius r, whose outer surface 29. A black body is at 727°C. It emits energy at a rate
radiates as a black body at a temperature T K is which is proportional to [AIPMT (Prelims)-2007]
given by (where  is Stefan’s constant) (1) (727)4 (2) (727)2
[AIPMT (Prelims)-2010] (3) (1000)4 (4) (1000)2
r 2T 4 r 2T 4 30. A black body at 1227°C emits radiations with
(1) (2)
R 2 4r 2 maximum intensity at a wavelength of 5000 Å. If the
temperature of the body is increased by 1000°C, the
r 4T 4 4r 2T 4 maximum intensity will be observed at :
(3) (4)
r4 R2
25. The two ends of a rod of length L and a uniform
cross-sectional area A are kept at two temperatures (1) 4000 Å (2) 5000 Å
dQ (3) 6000 Å (4) 3000 Å
T1 and T2 (T1 > T2). The rate of heat transfer, ,
dt 31. Which of the following circular rods, (given radius r
through the rod in a steady state is given by and length l) each made of the same material and
[AIPMT (Prelims)-2009] whose ends are maintained at the same
temperature will conduct most heat ?
dQ k (T1 – T2 ) dQ
(1)  (2)  kLA(T1 – T2 ) [AIPMT (Prelims)-2005]
dt LA dt
(1) r = 2r0; l = 2l0 (2) r = 2r0; l = l0
dQ kA(T1 – T2 ) dQ kL(T1 – T2 ) (3) r = r0; l = l0 (4) r = r0; l = 2l0
(3)  (4) 
dt L dt A Questions asked Prior to Medical Ent. Exams. 2005
26. A black body at 227°C radiates heat at the rate of 32. The coefficients of linear expansion of brass and
7 cals/cm2s. At a temperature of 727°C, the rate of steel are 1 and 2 respectively. When we take a
heat radiated in the same units will be brass rod of length l1 and steel rod of length l2 at
0°C, then difference in their lengths (l2 – l1) will
remain the same at all temperatures, if
(1) 50 (2) 112
(1) 12l1 = 22l2 (2) 1l2 = 2l1
(3) 80 (4) 60 (3) 1l1 = 2l2 (4) 1l22 = 2l12
33. The density of water at 20°C is 998 kg/m3 and at 39. Unit of Stefan’s constant is
40°C 992 kg/m 3 . The coefficient of volume (1) watt-m2-K4
expansion of water is
(2) watt-m2/K4
(1) 10–4/°C (2) 3 × 10–4/°C
(3) watt/m2-K
(3) 2 × 10–4/°C (4) 6 × 10–4/°C
(4) watt/m2K4
34. If 1 g of steam at 100°C steam is mixed with 1 g
of ice at 0°C, then resultant temperature of the 40. Consider a compound slab consisting of two
mixture is pieces of same length and different materials
having equal thicknesses and thermal conductivities
(1) 100°C (2) 230°C K and 2 K, respectively. The equivalent thermal
(3) 270°C (4) 50°C conductivity of the slab is
35. Heat is flowing through two cylindrical rods of the (1) 2/3 K
same material. The diameters of the rods are in the
ratio 1 : 2 and the lengths in the ratio 2 : 1. If the (2) 2 K
temperature difference between the ends is same,
then ratio of the rate of flow of heat through them (3) 3 K
will be (4) 4/3 K
(1) 2 : 1 (2) 8 : 1
41. Gravitational force is required for
(3) 1 : 1 (4) 1 : 8
(1) Stirring of liquid
36. A cylindrical rod has temperatures T1 and T2 at its
ends. The rate of flow of heat is Q (cal/s). If all the (2) Convection
linear dimensions are doubled keeping (3) Conduction
temperatures constant, then rate of flow of heat Q2
will be (4) Radiation
(1) 4Q1 (2) 2Q1 42. A black body is at a temperature of 500 K. It

emits energy at a rate which is proportional to
Q1 Q1
(3) (4) (1) (500)3
4 2
37. Two metal rods 1 and 2 of same lengths have same (2) (500)4
temperature difference between their ends. Their (3) 500
thermal conductivities are K1 and K2 and cross
sectional areas A1 and A2, respectively. If the rate (4) (500)2
of heat conduction in 1 is four times that in 2, then 43. Which of the following is closest to an ideal black
(1) K1A1 = K2A2 body?
(2) K1A1 = 4K2A2 (1) Black lamp
(3) K1A1 = 2K2A2 (2) Cavity maintained at constant temperature
(4) 4K1A1 = K2A2 (3) Platinum black
38. Consider two rods of same lengths and different (4) A lump of charcoal heated to high temperature
specific heats (S1, S2), conductivities (K1, K2) and
area of cross-sections (A1, A2) and both having 44. For a black body at temperature 727°C, its rate of
temperatures T1, and T2 at their ends. If rate of flow energy loss is 20 watt and temperature of surrounding
of heat due to conduction is equal, then is 227°C. If temperature of black body is changed to
(1) K1A1 = K2A2 1227°C then its rate of energy loss will be
K 1 A1 K 2 A2 (1) 304 W
(2) 
S1 S2 320
(2) W
(3) K2A1 = K1A2 3
K 2 A1 K 1 A2 (3) 240 W
(4) 
S2 S1 (4) 120 W
45. A beaker full of hot water is kept in a room. If it 51. A wire is stretched under a force. If the wire
cools from 80°C to 75°C in t1 minutes, from 75°C suddenly snaps, the temperature of the wire
to 70°C in t2 minutes and from 70°C to 65°C in t3 (1) Remains the same
minutes, then
(2) Decreases
(1) t1 < t2 < t3 (2) t1 > t2 > t3
(3) Increases
(3) t1 = t 2 = t 3 (4) t1 < t2 = t3
(4) First decreases then increases
46. The Wien’s displacement law expresses the 52. A uniform metal rod of 2 mm2 cross-section fixed
relation between at both ends is heated from 0°C to 20°C.
(1) Wavelength corresponding to maximum The coefficient of the linear expansion of the rod is
intensity and temperature 12 × 10–6/°C. Its Young’s modulus of elasticity is
1011 N m–2. The energy stored per unit volume of
(2) Radiation energy and wavelength the rod is
(3) Temperature and wavelength (1) 1440 J m–3
(4) Colour of light and temperature (2) 15750 J m–3
47. We consider the radiation emitted by the human (3) 1500 J m–3
body. Which one of the following statements is
correct? (4) 2880 J m–3
(1) The radiation emitted is in the infra-red region
SECTION - D
(2) The radiation is emitted only during the day
NEET Booster Questions
(3) The radiation is emitted during the summers
and absorbed during the winters 1. The approximate value of solar constant S
(in W/m2) is
(4) The radiation emitted lies in the ultraviolet
region and hence is not visible (1) 1260
48. If m denotes the wavelength at which the radiative (2) 1360

emission from a black body at a temperature T K (3) 1460
is maximum, then
(4) 1560
(1) m T 4
2. On the Fahrenheit scale, the temperature rise of
(2) m is independent of T nine degrees is equivalent to how much degree rise
in Celsius scale?
(3) m T
(1) 4
(4) m T–1
(2) 5
49. A black body has wavelength corresponding to
maximum intensity m at 2000 K. Its corresponding (3) 6
wavelength at 3000 K will be (4) 8
3 2 3. An aluminium scale [ = 2.4 × 10 –5 (C°) –1 ]
(1) m (2) m
2 3 calibrated at 25°C is measuring length of copper rod
[ = 1.7 × 10–5 (C°)–1] at 125°C. If the reading
16 81 shown is 100 m, then the correct reading is
(3) m (4) m
81 16
(1) 100.7 m (2) 100.07 m
50. The radiant energy from the sun, incident normally
(3) 107 m (4) 170 m
at the surface of earth is 20 kcal/m2 min. What
would have been the radiant energy, incident 4. If 150 gram ice at 0°C is mixed with 10 gram
normally on the earth, if the sun had a steam at 100°C then final temperature of
temperature, twice of the present one? composition under ideal condition is
(1) 320 kcal/m2 min (2) 40 kcal/m2 min (1) Less than 0°C (2) 0°C
(3) 160 kcal/m2 min (4) 80 kcal/m2 min (3) 10°C (4) 28°C
5. Four identical metal rods are connected as shown 10. A thin rod of negligible mass and area A suspended
in figure. Assume there is no heat loss through vertically from one end at °C. The rod is cooled
side walls of the rods and no radiation loss, the to 0°C but prevented from contracting by attaching
junction temperature in steady state is a mass at the lower end. If the Young’s modulus
of wire is Y and coefficient of linear expansion is
10°C , then the mass attached is
YA 2 YA
20°C 40°C (1) (2)
g g
YA YA
30°C (3)
g
(4)
g
(1) 22°C (2) 26°C
11. The steam point and the ice point of an arbitrary
(3) 25°C (4) 32°C temperature scale are marked as 80 S and 20 S.
6. Air outside a water lake is blowing at –40°C. It What will be the temperature in centigrade when
takes 10 hours for first two centimeter of ice layer this arbitrary scale reads 32 S?
to form over the lake. Time taken to form next four (1) 50°C (2) 40°C
centimeter of ice layer formation will be
(3) 30°C (4) 20°C
(1) 20 hours (2) 40 hours
12. The densities of wood and benzene at 0°C are 880
(3) 60 hours (4) 80 hours kg/m3 and 900 kg/m3 respectively. The coefficients
7. A hot solid metal ball of radius R is left for cooling of volume expansion are 1.2 × 10–3/°C for wood
in a room. Rate of cooling of the ball is proportional and 1.5 × 10–3/°C for benzene. The temperature at
which a piece of wood just sinks in benzene is
to
(1) 40°C (2) 70°C
(1) R2 (2) R
(3) 68°C (4) 80°C
1 1
(3) (4)
R R2 13. A cube and a sphere made of same material
shown in figure are allowed to cool under identical
8. Two different metal slabs of identical length are
conditions, then
joined in parallel. Their cross-sectional areas are A
and 2A. If their thermal conductivity coefficients are
k 1 and k 2 as shown then their equivalent
conductivity for one dimensional steady state heat 10 cm
transfer is
10 cm
Heat A (1) Cube will cool at a faster rate
k1
Flow 2A (2) Sphere will cool at a faster rate
k2
(3) Rate of cooling is independent of dimension of
k1  k2 body
(1) (2) k1  k2
2 (4) Both (2) and (3) are correct
k1  2k 2
(3) 2k1k2 (4) 14. A uniform thermometre scale is at steady state
3 with its 0 cm mark at 20°C and 100 cm mark at
9. A cup of tea cools from 70°C to 69.9°C in 6 100°C. Temperature of the 60 cm mark is
seconds. If temperature of surroundings is 30°C,
(1) 48°C
then time taken by the cup to cool from 50°C to
49.9°C is (assume Newton’s law of cooling) (2) 68°C
(1) 6 s (2) 9 s (3) 52°C
(3) 12 s (4) 15 s (4) 58°C
15. If Cp and Cv denote the specific heats (per unit 21. The molar specific heat at constant pressure of an
mass) of an ideal gas of molecular weight M, ideal gas is (7/2)R. The ratio of specific heat at
where R is the molar gas constant constant pressure to that at constant volume is
(1) Cp – Cv = R/M2 (2) Cp – Cv = R 9 7
(1) (2)
(3) Cp – Cv = R/M (4) Cp – Cv = MR 7 5
16. Two uniform rods AB and BC have Young’s modulii 8 5

(3) (4)
1.2 × 1011 N/m2 and 1.5 × 1011 N/m2 respectively. 7 7
If coefficient of linear expansion of AB is 22. 50 g ice at 0°C is dropped into a calorimeter
1.5 × 10–5/°C and both have equal area of cross containing 100 g water at 30°C. If thermal
section, then coefficient of linear expansion of BC, capacity of calorimeter is zero then amount of ice
for which there is no shift of the junction at all left in the mixture at equilibrium is
temperatures, is
(1) 12.5 g
C (2) 25 g
A B
(3) 20 g
(4) 10 g
(1) 1.5 × 10–5/°C (2) 1.2 × 10–5/°C 23. Heat energy at constant rate is given to two
substances P and Q. If variation of temperature (T)
(3) 0.6 × 10–5/°C (4) 0.75 × 10–5/°C
of substances with time (t) is as shown in figure,
17. Coefficient of linear expansion of a vessel then select the correct statement.
completely filled with Hg is 1 × 10–5/°C. If there is T
no overflow of Hg on heating the vessel, then P
coefficient of cubical expansion of Hg is Q
–5 –5
(1) 4 × 10 /°C (2) > 3 × 10 /°C
(3)  3 × 10–5/°C (4) Data is insufficient
t
18. A metallic tape gives correct value at 25ºC. (1) Specific heat of P is greater than Q
A piece of wood is being measured by this metallic
tape at 10ºC. The reading is 30 cm on the tape, (2) Specific heat of Q is greater than P
the real length of wooden piece must be (3) Both have same specific heat
(1) 30 cm (2) > 30 cm (4) Data is insufficient to predict it
(3) < 30 cm (4) Data is not sufficient 24. A bullet of mass 10 g moving with a speed of
20 m/s hits an ice block of mass 990 g kept on a
19. In a thermostat two metal strips are used, which
frictionless floor and gets stuck in it. How much ice
have different
will melt if 50% of the lost KE goes to ice? (initial
(1) Length temperature of the ice block and bullet = 0°C)
(2) Area of cross-section (1) 0.001 g
(3) Mass (2) 0.002 g
(4) Coefficient of linear expansion (3) 0.003 g
20. The coefficient of linear expansion of a crystalline (4) 0.004 g
substance in one direction is 2 × 10–4/°C and in 25. Heat is being supplied at a constant rate to
every direction perpendicular to it is 3 × 10–4/°C. the sphere of ice which is melting at the rate
The coefficient of cubical expansion of crystal is of 0.1 gm/s. It melts completely in 100 s. The rate
equal to of rise of temperature thereafter will be
(1) 5 × 10–4/°C (1) 0.4ºC/s
(2) 4 × 10–4/°C (2) 2.1ºC/s
(3) 8 × 10–4/°C (3) 3.2ºC/s
(4) 7 × 10–4/°C (4) 0.8ºC/s
26. In a calorimeter of water equivalent 20 g, water of 31. Three rods of same dimensions have thermal
mass 1.1 kg is taken at 288 K temperature. If conductivities 3K, 2K and K. They are arranged as
steam at temperature 373 K is passed through it shown, with their ends at 100°C, 50°C and 0°C.
and temperature of water increases by 6.5°C then The temperature of their junction is
the mass of steam condensed is
50°C
(1) 17.5 g 2K
3K
(2) 11.7 g 100°C
K
(3) 15.7 g 0°C
(1) 75°C
(4) 18.2 g
200
27. If the radius of a star is R and it acts as a black (2) °C
3
body, what would be the temperature of the star, in
which the rate of energy production is Q? (3) 40°C
( stands for Stefan's constant.) 100
(4) °C
1/ 4 3
 4R 2Q 
(1)   32. If wavelength of maximum intensity of radiation
   emitted by Sun and Moon are 0.5 × 10–6 m and
1/ 4
10 –4 m respectively, then the ratio of their
 Q  temperature is
(2)  2 
 4R  
1
(1)
Q 10
(3)
4R 2  1
(2)
1/ 2 50
 Q 
(4)  2 
 4R   (3) 100
28. Gravitational force is required for (4) 200

33. The three rods shown in figure have identical
(1) Stirring of liquid
dimensions. Heat flows from the hot end at a rate
(2) Convection of 40 W in the arrangement (a). Find the rates of
heat flow when the rods are joined as in
(3) Conduction arrangement (b). (Assume KAl = 200 W/m °C and
(4) Radiation KCu = 400 W/m °C)
29. Which of the following processes is reversible? 0°C Al Cu Al 100°C

(a)
(1) Transfer of heat by conduction
Al
(2) Transfer of heat by radiation 0°C 100°C
Cu
(3) Isothermal compression Al
(4) Electrical heating of a nichrome wire (b)

(1) 75 W (2) 200 W
30. Solar constant (S) depends upon the temperature
of the Sun (T) as (3) 400 W (4) 4 W
34. Two spheres of same material and radius r and 2r
(1) S  T are heated to same temperature and are kept in
(2) S  T 2 identical surroundings, ratio of their rate of loss of
heat is
(3) S  T 3 (1) 1 : 2 (2) 1 : 4
(4) S  T 4
(3) 1 : 6 (4) 1 : 8
35. If a graph is plotted by taking spectral emissive 40. A very thin metallic shell of radius r is heated to
power along y-axis and wavelength along x-axis temperature T and then allowed to cool. The rate
then the area below the graph above wavelength of cooling of shell is proportional to
axis is
1
(1) Emissivity (1) rT (2)
r
(2) Total intensity of radiation (3) r 2 (4) r 0
(3) Diffusivity 41. If an object at absolute temperature (T) radiates
(4) Solar constant energy at rate R, then select correct graph
36. A spherical black body with radius 12 cm radiates showing the variation of logeR with loge(T).
450 W power at 500 K. If the radius is halved and [Assume loge (e  A) > 0]
temperature is doubled, the power radiated in watt
would be loge(R)
(1) 225 (2) 450
(3) 900 (4) 1800
37. Three rods of same material, same area of cross- (1)
section but different lengths 10 cm, 20 cm and
30 cm are connected at a point as shown. What loge(T)
is temperature of junction O?
20ºC
logeR
20 cm
10 (2)
c m
cm
30 O
loge(T)
30ºC 10ºC
(1) 19.2ºC (2) 16.4ºC logeR
(3) 11.5ºC (4) 22ºC
1
38. If transmission power of a surface is , reflective
9 (3)
1 loge(T)
power is , then what is its absorptive power?
6
18 13 logeR
(1) (2)
13 18
3 15
(3) (4)
15 3 (4)
39. A solid cylinder of length L and radius r is heat
upto same temperature as that of a cube of edge logeT
length a. If both have same material, volume and
allowed to cool under similar conditions, then ratio 42. Two diagonally opposite corners of a square made of
of amount of radiations radiated will be (Neglect a four thin rods of same material, same dimensions
radiation emitted from flat surfaces of the cylinder) are at temperature 40°C and 10°C. If only heat
conduction takes place, then the temperature
a 2a difference between other two corners will be
(1) (2)
3r rL
(1) 0°C (2) 10°C
a2 a2
(3) (4) (3) 25°C (4) 15°C
rL 2rL
43. Bottom of a lake is at 0°C and atmospheric 45. Two bodies A and B of equal masses, area and
temperature is –20°C. If 1 cm ice is formed on the emissivity cooling under Newton’s law of cooling
surface in 24 h, then time taken to form next 1 cm from same temperature are represented by the
of ice is graph. If  is the instantaneous temperature of the
body and 0 is the temperature of surroundings,
(1) 24 h then relationship between their specific heats is
(2) 72 h log(– 0)
(3) 48 h
A
(4) 96 h B
44. The power received at distance d from a small t
metallic sphere of radius r(<<d) and at absolute (1) SA = SB (2) SA > SB
temperature T is P. If temperature is doubled and (3) SA < SB (4) None of these
distance reduced to half of initial value, then the
46. Assume that Solar constant is 1.4 kW/m2, radius
power received at that point will be
of sun is 7 × 105 km and the distance of earth from
(1) 4p centre of sun is 1.5 × 108 km. Stefan’s constant
is 5.67 × 10 –8 Wm –2 K –4, find the approximate
(2) 8p temperature of sun
(3) 32p (1) 5800 K (2) 16000 K
(4) 64p (3) 15500 K (4) 8000 K


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Chapter 11

Thermal Properties of Matter

z Heat Transfer MEASUREMENT OF TEMPERATURE

0°C 273 K 32°F

Measured temperature – LFP

C 0 K  273 F  32 Measured temprature – LFP

Solution : Let that temperature be x, then

So, at – 40°C, Fahrenheit reading is also – 40°F.

Solution : Upper fixed point = –60 °X

x  ( 180.5) 350  273.15

The pressure and temperature are related as Pt = P0(1 + t )

then the resistance and temperature are related as

These thermometers can measure temperature from – 182°C to 1200°C.

4. Thermoelectric Thermometers : It is based on Seebeck effect. In this thermometer two distinct

7. Magnetic Thermometers : Temperatures below 1 K are measured by ‘magnetic thermometers’ which

(i) Temperature coefficient of resistance of platinum,

(ii) Value of temperature t.

Solution : (i) Temperature coefficient of resistance is given by

R100  R0 4.71  3.70

(ii) For temperature t, we have

5.29  3.70 1.59

z Thermal Expansion is of Three Types

If a body has length L0 at temperature T0, then its length L at a temperature T = T0 + T is

The unit of  is K–1 or (C°)–1.

The value of  is of the order of 10–6 to 10–5 per kelvin

z Relation Between Linear Expansion, Superficial expansion and Volume Expansion

z Apparent Expansion of Liquids

There are two coefficients of expansion in case of liquids.

z Change of Density with Temperature

L = L0T = (1.2 × 10–5 K–1) (50 m) (15 K) = 9.0 × 10–3 m = 9.0 mm

L = L0 + L = 50.000 m + 0.009 m = 50.009 m

Thus, the length at 35°C is 50.009 m.

Vmercury – Vglass = 2.7 cm3

(3) 15°C (4) 12.5°C

Value of c depends on material of body and temperature range.

The SI unit of C is J/mol K.

Note : (1) Specific heat of water is

sw  4186 J kg1K 1  1 calg1 (C)1 .

Fig.: Graph of temperature vs time for a specimen

Q4 = msswt = 10 g × 1 cal/g°C × 100 = 1000 cal

Heat supplied to steam

Maximum heat absorbed by ice 5400 cal

 m.SA (30°C – 26°C) = m.SB(26°C – 20°C)

Fig.: Phase diagram for water

(3) – 5°C (4) 0°C

(1) 1:2 (2) 1:8

(3) 1:4 (4) 2:1

(3) 1600 gm (4) 150 gm

13. Work done in converting 1g of ice at –10°C into steam at 100°C is

(1) 3.04 kJ (2) 6.05 kJ

(3) 0.721 kJ (4) 0.616 kJ

(1) 540 cal/g (2) 536 cal/g

(3) 270 cal/g (4) 480 cal/g

(1) 20 gm of water (2) 20 gm of ice

(3) 10 gm ice and 10 gm of water (4) 5 gm ice and 15 gm water

(1) 0.96°C (2) 1.02°C

(3) 1.16°C (4) 0.23°C

(1) 4800 W (2) 336 W

(3) 1.33 W (4) 0.75 W