OC 450 Climatic Extremes Problem #1 Due: Friday, Jan 16

1. The steady-state surface temperature of the earth can be estimated from a simple heat budget where
it is assumed that there is a balance between heat input to the surface via incoming short wave (SW)
solar radiation and heat loss from the surface via outgoing long wave (LW) back radiation. This heat
balance is described by the following relationship.
I (1 - ) = f T4,
where I = solar insolation at 342 W/m2 (W= watts), represents global albedo (reflectivity), f is
atmospheric transmissivity to LW radiation, which varies inversely with greenhouse gas
concentrations, is the Stefan-Boltzman constant [5.67x10-8 W/(m2 K4)] and T is absolute
temperature (K).

a. Calculate the surface temperature for the earth assuming there is no refection of insolation to
space (global albedo is zero) and the atmosphere is transparent to LW radiation (no LW
adsorption in atmosphere).
a. Ans. T=278.7 K or 5.7 C
b. Calculate the surface temperature for the earth assuming that clouds and earths surface
reflect 30% of incoming insolation (global albedo) and the atmosphere is transparent to LW
radiation.
a. Ans. T=254.9 K or 18.1 C
c. Calculate the surface temperature for the earth assuming a global albedo of 30% and an
atmospheric composition that transmits 61% of surface back radiation into space (current
situation on earth).
a. Ans. T=288.4 K or 15.4 C
d. Calculate the surface temperature at the end of this century assuming the projected increase in
the atmospheric CO2 concentration by 2100 decreases current LW transmissivity by 2.5%.
a. Ans. T=290.2 K or 17.2 C (note: a 2.5% reduction in f was meant to imply that f =
0.61*0.975 = 0.595)
e. Calculate the surface temperature at the end of this century assuming that in 2100 reduced sea
and land ice coverage has decreased the earths current albedo by 3% in addition to the CO2
effect determined in part d.
a. Ans. T=291.3 K or 18.3 C (note: a 3% reduction in was meant to imply that =
0.3*0.97 = 0.291)
f. Explain why the above scenario e, when compared to scenario d, is an example of a
positive or negative feedback.
Ans. Positive Feedback. The temperature change rate in e exceeds that in d. The likely
reason is that a temperature increase melts ice, which in turn reduces albedo, increases SW
insolation reaching the earths surface and accelerating temperature increase

2. At some altitude in the atmosphere there exists a steady-state heat flux balance between net solar
insolation flux towards earth [I*(1-), with =0.30] and long wave back radiation heat flux away from
earth (fT4), assuming at above this altitude the atmosphere is essentially transparent to LW radiation
(f=1). Calculate the height in the atmosphere where this heat flux balance exists for three scenarios.

Hint: You will need to use the adiabatic lapse rate (cooling rate of air per change in altitude) =
6.5C/km to answer the question.

Ans. The temperature at which there is a balance between incoming SW and outgoing long wave is
254.9 K, calculated from T = [342*(1- 0.3)/(1*5.67x10-8)]0.25. The altitude of the surface where this
heat flux balances exists is calculated from the difference between the temperature on this surface
(254.9K) and the temperature on the surface of the earth and the adiabatic cooling rate. That is, the
altitude equals the temperature difference (or gradient) divided by the lapse rate. (Realize that a
temperature difference is equal in magnitude whether expressed in C or K, thus the adiabatic cooling
rate = 6.5 C or K per kilometer.)

a. First, for the earths heat budget under a transparent atmosphere (scenario b in Question 1).

ans. Temperature at the surface of the earth is 254.9 under transparent atmospheric conditions, which is
the same as the temperature needed for a heat flux balance. So the altitude equals 0 km, i.e., the surface
of the earth (atlitude = Temp / lapse rate = (254.9-254.9 K)/ (6.5 km/ K) = 0 km.

b. Second, for the earths current heat budget (scenario c in Question 1).

ans. Current Conditions: Altitude = (Surface temperature 254.9 K)/ (6.5 K/km)
= (288.4 254.9 K)/(6.5 K/km)
= 5.2 km

Third, for the predicted heat budget in 2100 (scenario e in Question 1).

Future Conditions: Altitude = (Surface temperature 254.9 K)/ (6.5 K/km)

What happens to the height of this heat flux balance surface in the atmosphere as the level of
greenhouse gases in the atmosphere increases?

Ans. As GHGs increase and transmissivity decreases, the altitude of the surface where there is a heat
flux balance (T=254.9K) will increase.

3. The rate of temperature change of a reservoir is related to the difference between heat input and
heat loss and the total heat content of the reservoir and can be expressed by the following
relationship.

Heat / time = Heat Input Heat Loss,

The heat content of a volume of seawater equals V**c*T, where V=volume (m3), = density
(kg/m3), c = specific heat (Joules/[kg K]), T = temperature (K).

The average heat input rate at the earths surface equals 47% of the incoming solar insolation at the
top of the atmosphere (342 W/m2), see Fig. 2-3 in Text. (Assume the density and specific heat of
seawater are 1024 kg/m3 and 3985 Joules/(kg K), respectively, and 1Watt = 1 Joule/sec)

a. Calculate the temperature increase for a 100m deep layer of the ocean receiving average solar
insolation for six months assuming there is no heat loss from the ocean.

ans. (V**c*T)/t = Heat Input Heat Output

Heat Input = 342 W/m2 *0.47 = 160.7 W/m2 = 160.7 Joules/(m2 sec) (Note: 1Watt = 1 Joule/sec)
Heat Output = 0 (assumed)
The time rate of change of temperature (Temp/time) =

T/t = Heat Input/ (V**c) = Surface Area* Heating Rate/ (V**c)

Since heat input equals surface area * heating rate and volume equals surface area * depth, then

T/t = Heating Rate/ (Z**c)

Heat capacity = 3985 J/(kg K)

T/t = (342 (J/s)/m2 *0.47) / (100m*1024 kg/m3*3985J/(kg K)

Over six months, time = 1.58x107 sec

Thus over six months the change in temperature is:

Temperature = 3.94x10-7 K/s * 1.58x107 sec

b. Repeat the calculation of the temperature increase for a 100m deep layer of the ocean receiving
average solar insolation for six months but assume there is heat loss via LW back radiation from the
surface ocean at a mean temperature of 15C.

ans. (V**c*T)/t = Heat Input Heat Output

Heat Input = 342 W/m2 *0.47 = 160.7 W/m2 = 160.7 Joules/(m2 sec) (Note: 1Watt = 1 Joule/sec)

Heat Output = f * *T(K)4

(V**c*T)/t = 160.7 390.1 = -229.4 W/m2

The time rate of change of temperature

T/t = (-229.4 J/m2 s) / (100m*1024 kg/m3*3985J/(kg K)

Over six months, time = 1.58x107 sec

Thus over six months the change in temperature is:

Temperature = -5.62x10-7 K/s * 1.58x107 sec

Ans. The calculated temperatures changes are of the same order as the sea surface temperature
(SST) changes presented in Fig 2-10.

d. What additional heat flux is missing from these calculations that helps explain the difference
between the heating rates calculated in parts a and b? (Hint: Look at Fig. 2.3). Estimate the
magnitude of this missing heat flux.

Ans. The missing heat flux is that associated with the downward radiation of LW heat from the
clouds to the earths surface, which comprises about 96% of the LW radiation lost from the surface
layer of the earth.

e. Explain why the winter-summer heating rate over land is much greater than over the ocean (see Fig
2-10).

Ans. The specific heat of seawater (4000J/kg/K) exceeds that for soils and rocks (~800 J/kg/K)
but the density of rocks is almost 3x the density of seawater which offsets some of the difference
in specific heat. A more important effect on the heat capacity difference between ocean and land
occurs because the surface ocean is mixed downward by winds to a depth of ~100m. On land, in
contrast, the depth of the soil reservoir influenced by seasonal heating is much shallower (<1 m)
and, thus, the total heat capacity is much smaller in soils than for the ocean. Since temperature
increase rate is inversely proportional to total heat capacity (T/t = Heating rate
/(Volume*density*heat capacity), the SST will increase less than soil temperature for the same
heating rate because the volume (penetration depth of heat) is much greater.

4. To demonstrate the relationship between residence time and response time, well calculate the time
rate of change in the content of a reservoir that responds to a change in input rate. The equation that
describes the time rate of change of the content of the reservoir is the following:

Mt = Mf (Mf - Mo)*e(-t/tau)

Where t = time, tau = residence time, Mo is the initial content under the initial input rate and Mf is the
final content of the reservoir under the new input rate. This is a general equation which can represent
the change in the heat content of the atmosphere, the water content of a bathtub, the carbon content of
the terrestrial biosphere, etc.

Assume Mf = 100 and Mo = 0 and tau = 10 days.

a. Calculate content of the reservoir (Mt) at 5 day intervals between 0 and 60 days.
See excel output below.
b. Repeat the calculation in part a, when tau = 20 days (same Mf and Mo as in part a).
See excel output below.
c. Repeat the calculation in part a, when tau = 60 days (same Mf and Mo as in part a).
See excel output below.
d. Plot the results, i.e., Mt vs t (days), for parts a, b and c on the same figure.
See excel output below.

e. How does the shape of the three curves compare to the one describing the change in
temperature of the flask in Fig1-6 in the text?
 In general the shape of the Mt vs time curves are similar to the curve shape of the time rate of
change of heat content shown in Fig 1-6. This demonstrates that this very simple equation
yields a reasonable approximation to real system.

f. Calculate the time it takes to attain 95% of the final content (Mf) in parts a and b. How does
this time compare to the residence time in each case?
g.
When Tau= 10 days, Time=30days. When Tau= 20 days, Time=60days
In each case the time to reach 95% of the final value (Mf) is three times the residence time.

tau= 20 10 60
Mf 100
Mo 0
Time(days) Mt
0 0.0 0.0 0.0
5 22.1 39.3 8.0
10 39.3 63.2 15.4
15 52.8 77.7 22.1
20 63.2 86.5 28.3
25 71.3 91.8 34.1
30 77.7 95.0 39.3
35 82.6 97.0 44.2
40 86.5 98.2 48.7
45 89.5 98.9 52.8
50 91.8 99.3 56.5
55 93.6 99.6 60.0
60 95.0 99.8 63.2