Temperature and Heat: Leaving Cert Physics Long Questions 2017 - 2002
Temperature and Heat: Leaving Cert Physics Long Questions 2017 - 2002
Temperature and Heat: Leaving Cert Physics Long Questions 2017 - 2002
Contents
Temperature: ordinary level questions ........................................................................................................................... 2
Temperature: higher level questions .............................................................................................................................. 3
Heat: ordinary level theory questions ............................................................................................................................. 5
Heat: ordinary level maths questions.............................................................................................................................. 7
Heat: higher level maths questions ................................................................................................................................. 9
Solutions to ordinary level maths questions ................................................................................................................. 13
Solutions to all higher level questions........................................................................................................................... 15
1
Temperature: ordinary level questions
2014 Question 7 (a) [Ordinary Level]
The temperature of an object can be measured using a thermometer which is based on a suitable
thermometric property.
(i) What is heat?
(ii) What is meant by temperature?
(iii)Give an example of a thermometric property.
(iv) The SI unit of temperature is the kelvin. Give another temperature scale.
(v) Express 310 K in the units of the scale you have named in part (iv).
2
Temperature: higher level questions
2003 Question 12 (b) [Higher Level]
(i) What is the difference between heat and temperature?
(ii) The emf of a thermocouple can be used as a thermometric property.
Explain the underlined terms.
(iii)Name a thermometric property other than emf.
(iv) Explain why it is necessary to have a standard thermometer.
3
2013 Question 12 (d) [Higher Level]
(i) What is meant by the term thermometric property?
This graph was obtained during an experiment where the resistance R of a thermistor was measured as
its temperature θ was raised from 0 °C to 100 °C (as measured by a mercury-in-glass thermometer).
The thermistor is used in a circuit to keep the water in a tank at a constant temperature.
(ii) What is the temperature of the water when the resistance of the thermistor is 420 kΩ?
(iii)A thermocouple thermometer has emf values of 0 μV at 0 °C and 815 μV at 100 °C.
When the thermocouple thermometer was placed in the tank of water, its emf was found to be 319 μV.
What is the temperature of the water in the tank as measured by the thermocouple thermometer?
(iv) Why do the thermistor and the thermocouple thermometer give different temperature readings for the
water in the tank?
4
Heat: ordinary level theory questions
2014 Question 7 (b) [Ordinary Level]
The photograph shows an experiment to compare the heat transfer in different metals.
A piece of wood is placed in a drop of wax at the end of each piece of metal and a heat
source is used to heat the metals at the centre of the apparatus.
(i) How is heat transferred in metals?
(ii) Name the two other methods of heat transfer.
(iii)How can this experiment be used to find out which of the metals is best at allowing
heat transfer?
(iv) State one way to make sure that this is a fair test.
5
2006 Question 7 (a) [Ordinary Level]
(i) Heat can be transferred in a room by convection.
(ii) What is convection?
(iii)Name two other ways of transferring heat.
(iv) Describe an experiment to demonstrate convection in a liquid.
(v) In an electric storage heater, bricks with a high specific heat capacity are heated overnight by passing an
electric current through a heating coil in the bricks.
The bricks are surrounded by insulation.
Why is insulation used to surround the bricks?
(vi) Name a material that could be used as insulation.
(vii) Explain how the storage heater heats the air in a room.
6
Heat: ordinary level maths questions
2013 Question 9 (b) [Ordinary Level]
A kettle contains 1.3 kg of water with a specific heat capacity of 4200 J kg−1 K–1.
The temperature of the water rises from 10 °C to 80 °C during a three-minute period.
(i) Calculate the energy gained by the water.
(ii) Calculate the power rating of the kettle, assuming all of the electrical energy is used to heat the water.
The diagram shows a kettle which is filled with 500 g of water which is initially at a
temperature of 20 °C.
The heating element of the kettle has a power rating of 0.8 kW.
We assume all the heat is transferred to the water.
(ii) Find the energy required to raise the temperature of the water to 100 °C.
(iii)What is the energy supplied by the element per second?
(iv) How long will it take the kettle to heat the water to 100 °C?
(v) Why are handles of kettles often made of plastic?
(vi) How is the heat transferred throughout the liquid in the kettle?
(vii) Why is the heating element of a kettle made of metal?
(viii) The heat source for a kettle is placed at the bottom. Suggest why this is the case.
(specific heat capacity of water = 4180 J kg−1 K−1)
7
2006 Question 7 (b) [Ordinary Level]
The total mass of the bricks in a storage heater is 80 kg and their specific heat capacity is 1500 J kg–1 K–1.
During a ten-hour period the temperature of the bricks rose from 15 oC to 300 oC.
(i) Calculate the energy gained by the bricks;
(ii) Calculate the power of the heating coil.
8
Heat: higher level maths questions
2006 Question 12 (c) [Higher Level]
(i) Define power.
(ii) Define specific heat capacity.
(iii)400 g of water at a temperature of 15 oC is placed in an electric kettle.
The power rating of the kettle is 3.0 kW.
Calculate the energy required to raise the temperature of the water to 100 oC.
(iv) Calculate the energy supplied by the kettle per second.
(v) Calculate the least amount of time it would take to heat the water to 100 oC.
(vi) In reality, the time taken to heat the water will be greater. Explain why.
(specific heat capacity of water = 4200 J kg–1 K–1)
(cmilk = 3.90 × 103 J kg–1 K–1, cwater = 4.18 × 103 J kg–1 K–1, chot drink = 4.05 × 103 J kg–1 K–1
specific latent heat of vaporisation of water = 2.34 × 106 J kg–1)
9
2004 Question 7 [Higher Level]
(i) Define specific heat capacity.
(ii) Define specific latent heat.
(iii)500 g of water at a temperature of 15 0C is placed in a freezer.
The freezer has a power rating of 100 W and is 80% efficient.
Calculate the energy required to convert the water into ice at a temperature of –20 oC.
(iv) How much energy is removed every second from the air in the freezer?
(v) How long will it take the water to reach a temperature of –20 oC?
(vi) Allowing a liquid to evaporate in a closed pipe inside the freezer cools the air in the freezer. The vapour
is then pumped through the pipe to the outside of the freezer, where it condenses again.
Explain how this process cools the air in the freezer.
(vii) The freezer causes the room temperature to rise. Explain why.
(iii)As part of his presentation, Joule proposed that the temperature of the water at the bottom of the
Niagara Falls would be 0.12 °C greater than that at the top, due to gravitational potential energy being
converted into heat energy.
Calculate the height of the Niagara Falls.
(v) In a heat pump, a fluid is used to transfer energy from a cold body to a warmer body.
Describe the operation of a heat pump and explain how a heat pump can be used to reduce the
temperature of a cold region, for example the interior of a refrigerator.
(vi) State two desirable physical properties of the fluid used in a heat pump.
The fluid in the heat pump of a refrigerator has a specific latent heat of vaporisation of 4.6 MJ kg–1.
The internal volume of the refrigerator is 0.6 m3.
The heat pump removes 12 kJ of energy from the air in the refrigerator as the fluid evaporates.
(vii) Calculate the mass of fluid that has evaporated
(viii) Calculate the fall in temperature of the air in the refrigerator.
(specific heat capacity of water = 4200 J kg–1 K–1; acceleration due to gravity = 9.8 m s–2;
density of air = 1.23 kg m–3; specific heat capacity of air = 1005 J kg–1 K–1)
10
2012 Question 12 (c) [Higher Level]
The graph shows the variation in temperature θ of 150 g of crushed ice when it was supplied with energy ΔE
at a constant rate.
11
2009 Question 11[Higher Level]
Read the following passage and answer the accompanying questions.
The sun is a major source of ‘green’ energy. In Ireland solar heating systems and geothermal systems are
used to get energy from the sun.
There are two main types of solar heating systems, flat-plate collectors and vacuum-tube collectors.
1.
A flat-plate collector is usually an aluminium box with a glass cover on top and a blackened plate on the
bottom. A copper pipe is laid on the bottom of the box, like a hose on the ground; water is passed through
the pipe and transfers the absorbed heat to the domestic hot water system.
2.
In a vacuum-tube collector, each tube consists of an evacuated double-walled silvered glass tube in which
there is a hollow copper pipe containing a liquid. The liquid inside the copper pipe is vaporised and expands
into the heat tip. There the vapour liquefies and the latent heat released is transferred, using a heat
exchanger, to the domestic hot water system. The condensed liquid returns to the copper pipe and the cycle
is repeated.
In a geothermal heating system a heat pump is used to extract solar energy stored in the ground and transfer
it to the domestic hot water system.
(a) What is the maximum energy that can fall on an area of 8 m2 in one hour if the solar constant is 1350 W
m–2?
(b) Why is the bottom of a flat-plate collector blackened?
(c) How much energy is required to raise the temperature of 500 litres of water from 20 0C to 50 0C?
(d) The liquid in a vacuum-tube solar collector has a large specific latent heat of vaporisation. Explain why.
(e) Name the three ways that heat could be lost from a vacuum-tube solar collector.
(f) How is the sun’s energy trapped in a vacuum-tube solar collector?
(g) Describe, in terms of heat transfer, the operation of a heat pump.
(h) Give an advantage of a geothermal heating system over a solar heating system.
(specific heat capacity of water = 4200 J kg–1 K–1; density of water = 1000 kg m–3; 1 litre = 10–3 m3)
12
Solutions to ordinary level maths questions
2017 Question 12 (b)
(i) Explain the underlined terms.
heat pump: means of transferring heat from a cold region to a warm region
specific latent heat : heat needed to change the state of 1 kg of a substance
2016 Question 9
(i) What is meant by latent heat?
Latent heat is the heat energy required to change the state of a substance (without changing temperature)
(ii) Name an instrument used to measure temperature.
Thermometer
(iii)Calculate the energy removed from the water to reduce its temperature to 0C
Q = mcΔθ = 0.75 × 4200 × 20 = 63000 J
(iv) Calculate the energy removed from the water to convert the water at 0C to ice at 0C
Q = ml = 0.75 × (3.3 × 105) = 247500 J
(v) Calculate the energy removed from the water to cool the ice at 0C to ice at −15 0C.
Q = mcΔθ = 0.75 × 2200 × 15 = 24750 J
(vi) How long will it take for the freezer to remove 9000 J of energy from the water?
𝑊𝑜𝑟𝑘 𝑊𝑜𝑟𝑘 9000
𝑃 = 𝑡𝑖𝑚𝑒 𝑡𝑖𝑚𝑒 = 𝑝𝑜𝑤𝑒𝑟 𝑡𝑖𝑚𝑒 = 300 time = 30 s
(vii) As the water freezes, the glass bottle cracks and shatters. Explain why this occurs.
The water expands (and the glass contracts).
2015 Question 9
(ii) E = mcΔθ = (5×4180×80) = 167 200 J
(iii) The energy is supplied by the kettle which has a power rating of 0.8 kW = 800 Watts which corresponds
to 800 Joules per second.
(iv) 167 200 Joules are required to heat the water, and energy is supplied at a rate of 800 Joules per second
So time taken is 167200/800
= 209 s
2013 Question 9
(i) E = mcΔθ = (1.3)(4200)(80-10) = 3.8 ×105 J
(ii) P = Energy/time = 3.8 ×105 / 180 = 2123.3 W
2012 Question 9
(ii) t = T - 273(.15)
(i) E = ml = (20 ×10-3)(3.34 ×105) = 6.68 ×103 J
(ii) E = mcΔθ = (20×10-3)(4.18 ×103)(5) = ) 418 J
13
2008 Question 7
(i) 100 – 20 = 80 °C
(ii) Q = m cΔθ = 0.5 × 4200 × 80 = 168 000 J
(iii) 2 kW = 2,000 W = 2,000 J per second.
(iv) P = W/t so t = W/P t = 168 000/2,000 = 84 secs.
2006 Question 7
(i) Q = mcΔθ Q = (80)(1500)(285) = 34 200 000 J = 4.2 MJ
(ii) P = W/ t P = 34 200 000 / (10×60×60) = 950 W
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Solutions to all higher level questions
2016 Question 7
𝐸 12 ×103
𝑚= m= = 0.0026 kg
𝑙 4.6 ×106
The heat pump removes 12 kJ of energy from the air in the refrigerator as the fluid evaporates.
So Energy removed = 12000 J.
cair = 1005 J kg–1 K–1
12000
Δθ =
(0.74)(1005)
Δθ = 16.1 °C
15
2015 Question 12 (c)
(i) Explain the underlined terms.
Thermometric property: one that changes (measurably) with temperature
Emf: A voltage when applied to a full circuit is called an emf (electromotive force).
16
2014 Question 12 (c)
(i) Define specific latent heat.
This is the heat needed to change the state of 1 kg of a substance without a change in temperature.
(iii)Calculate the minimum temperature of the water when the ice has melted.
Heat lost: the heat lost by the water = mcΔθ3, where Δθ3 is the difference between the initial temperature of
the water (which was 24 °C) and the final temperature of the system T.
So Δθ = (24 – T)
2281.125T = 34357.5
T = 15.06 °C
(ii) What is the temperature of the water when the resistance of the thermistor is 420 kΩ?
.Read directly from the graph to get approximately 370 C
(iii)What is the temperature of the water in the tank as measured by the thermocouple thermometer?
As the temperature goes from 0C to 100 0C, the emf goes from 0 V to 815 μV
100
So an increase of 1μV corresponds to a temperature difference of 815 or 0.1227 0C
100
319 μV therefore corresponds to a temperature of 815 × 319 = 39.14 0C
(iv) Why do the thermistor and the thermocouple thermometer give different temperature readings
for the water in the tank?
Each of the thermometers has a different thermometric property which changes differently with
temperature
17
2012 Question 12 (c)
(i) Explain the shape of the graph.
Temperature of ice increased from –3o to 0o as energy is added.
Ice temperature stays at 0 0C while ice is melting / changing state - the heat taken in at this stage is
known as latent heat.
Once all the ice has melted the water temperature increases to 1oC.
(ii) Describe how energy could have been supplied at a constant rate.
Using a heating coil (ideally with a joulemeter and rheostat to monitor and adjust the energy supplied).
(iii)Using the graph, estimate the specific latent heat of fusion of ice.
From the graph we can see that the energy required to melt 0.15 kg of ice = (59 kJ – 10 kJ) = 49 kJ
Q = ml
𝐸 49 ×102
𝑙= = l = 3.27 × 105 J Kg-1
𝑚 0.15
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2011 Question 7 (a)
(a)
(i) If the final temperature of the drink is to be 70 °C, what mass of steam should be added?
Energy gained by the milk = energy lost by the steam when condensing + energy lost by this condensed
water cooling down
31.2×103
31.2 × 103 = (2.4654 × 106 )𝑚𝑠𝑡𝑒𝑎𝑚 𝑚𝑠𝑡𝑒𝑎𝑚 = 2.4654×106 = 12.655×10-3 kg = 12.66 g
(b)
How is the energy lost by each of these processes reduced for a hot drink supplied in a disposable
cup?
Conduction – The material the cup is made from is a good insulator
Evaporation – use a lid
Convection – Use a lid /insulation
(c)
(i) How would you demonstrate the principle of operation of a thermocouple?
A thermocouple is comprised of two different metals joined
together at both ends to form two junctions.
One junction is connected to a liquid of known temperature
like ice-water which would be at 0 0C; this is the cold or
reference junction.
The other junction is connected to the liquid whose
temperature is to be measured.
An emf /voltage is produced (as seen in the diagram).
Increase the temperature of the heat source and note the new
temperature (using an existing thermometer) and new voltage.
Plot a graph of emf against temperature.
You can now use the graph with this thermocouple to establish
the temperature of the liquid in the future (dip the measuring junction into the liquid note the voltage and
use the graph to get the corresponding temperature).
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2009 Question 11
(a) What is the maximum energy that can fall on an area of 8 m2 in one hour if the solar constant is
1350 W m–2?
1350 joules of energy fall on one m2 every second.
So the amount of energy falling on 8 m2 in one hour corresponds to 1350 × 8 × 60 × 60 = 3.9 × 107 J
(c) How much energy is required to raise the temperature of 500 litres of water from 20 0C to 50 0C?
{There are one thousand litres in one cubic metre, so 1 litre = 1 × 10–3 m}
𝑚𝑎𝑠𝑠
𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 𝑣𝑜𝑙𝑢𝑚𝑒 mass = (density)(volume) mass = (1000)(500 × 10–3) = 500 kg.
Q = mc = (500)(4200)(30) = 6.3 × 107 J
(d) The liquid in a vacuum-tube solar collector has a large specific latent heat of vaporisation. Explain
why.
So that the liquid can absorb a lot of energy per kg in the heat exchanger during a change of state.
(e) Name the three ways that heat could be lost from a vacuum-tube solar collector.
Conduction, convection, radiation
(h) Give an advantage of a geothermal heating system over a solar heating system.
Geothermal system functions all the time whereas a solar heating system works only during sunshine.
(iii)Calculate the energy required to raise the temperature of the water to 100 oC.
Q = mcΔθ
Q = (0.40)(4200)(85) = 1.428 x 105 J
(v) Calculate the least amount of time it would take to heat the water to 100 oC.
𝐸𝑛𝑒𝑟𝑔𝑦 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑡𝑜 ℎ𝑒𝑎𝑡 𝑡ℎ𝑒 𝑤𝑎𝑡𝑒𝑟 1.428×105
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 = 𝑡𝑖𝑚𝑒 = time = 47.6 seconds
𝑟𝑎𝑡𝑒 𝑎𝑡 𝑤ℎ𝑖𝑐ℎ 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑠 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 3000
(vi) In reality, the time taken to heat the water will be greater. Explain why.
Energy will be lost to the surroundings.
20
2004 Question 7
(i) Define specific heat capacity
The specific heat capacity of a substance is the heat energy needed to change one kilogram of the
substance by one Kelvin.
(iii)Calculate the energy required to convert the water into ice at a temperature of –20 oC.
There are 3 separate stages here:
Cooling from 15 0C to 0 0C: Q = mcΔθ = (0.5)(4200)(15) = 31500 J
Change of state: Q = ml = (0.5)(3.3 × 105) =165000 J
Cooling ice from 0 oC to -20 oC: Q = mcΔθ = (0.5)(2100)(20) =21000 J
5
Total energy required = Qt = Q1 +Q2 +Q3 =217500 = 2.2 × 10 J
(iv) How much energy is removed every second from the air in the freezer?
{Power rating = 100W = 100 Joules per second.
But 80% efficiency means that the useful power is actually 80 W}
So 80 J of energy is removed every second.
(v) How long will it take the water to reach a temperature of –20 oC?
𝑒𝑛𝑒𝑟𝑔𝑦 𝑤ℎ𝑖𝑐ℎ 𝑛𝑒𝑒𝑑𝑠 𝑡𝑜 𝑏𝑒 𝑟𝑒𝑚𝑜𝑣𝑒𝑑 217500
𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑗𝑜𝑢𝑙𝑒𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 𝑟𝑒𝑚𝑜𝑣𝑒𝑑 𝑝𝑒𝑟 𝑠𝑒𝑐𝑜𝑛𝑑 𝑡 = 80 = 2700 s
(vi) Explain how this process cools the air in the freezer.
This change of state which takes place inside the pipe requires energy (latent heat). This energy is
available from the air surrounding the pipe which is inside the freezer; this therefore lowers the
temperature of the freezer.
(vii) The freezer causes the room temperature to rise. Explain why.
When the vapour condenses inside the pipe (at the back of the freezer) latent heat energy is released to
the surroundings. This causes the air outside the pipe (the air in the room) to increase in temperature.
21