Zonal Energy Balance Model Practical : Answers
1. Set the model parameters for Runs 1 and 2 to the default parameters given in table 1.
a) where is the final ice-edge?
At about 70o latitude
b) what are the mean temperature and mean albedo?
<T> = 14.9 oC , < > = 0.31
c) what are the approximate temperatures at the pole and the equator?
-15 oC and 25 oC
d) does this look like the real Earth?
Yes, sort of !
2. How do you expect the ice-edge, mean albedo and mean temperature to vary with the parameter
K? Run the model with different values of K to check your intuition.
Increased K  increased heat transport reduced ice-sheetreduced albedoincreased T
a) what is the mean temperature and albedo in the absence of heat transport?
<T> = 9.0 oC , < > = 0.35
b) at what K value does the model become completely ice-free?
K = 3.77 W m-2 oC-1
c) what are the implications of this sensitivity to K with regard to variations in poleward heat
transports (e.g. variations in the Gulf stream)? Is this realistic?
The model climate is extremely sensitive to heat transport such that a small increase in the
thermohaline circulation could completely melt the ice caps. This sensitivity is probably
unrealistic given the known variability in the Earth’s heat transports.
3. Return parameters to their default values in table 1.
a) By reducing the “Fraction of Solar Constant” in Run 1 determine the value at which the
Earth becomes completed ice-covered.
fS = 0.83
b) Set the initial ice-edge to 0o (i.e. “Snowball Earth'”) and increase the solar constant from the
value found in (a). At which value does the ice-sheet completely melt?
fS = 1.17
c) Set the initial ice-edge to 90o (i.e. ice-free Earth) and reduce the solar constant from the
value found in (b). At which value does the ice begin to reappear?
fS = 0.99
d) Sketch phase diagrams of mean albedo versus fractional solar constant, and mean
temperature versus fractional solar constant. Initialise Run 1 and Run 2 with ice-edges at 0o
and 90o respectively, and increase the solar constant values from 0.8 to 1.2 in steps of 0.05.
Fill in the mean values of temperature and albedo in the workspace table at the bottom of
the “Graphs” sheet. This will produce phase space diagrams below the table.
50
40
30
20
10
0
-10 0
-20
-30
-40
-50
EBM Climate Equilibria
0.7
0.6
Run 1
0.5
1
1.5
Run 2
Mean Albedo
Mean Temperature (deg C)
EBM Climate Equilibria
0.5
0.4
Run 1
0.3
Run 2
0.2
0.1
0
0
Fraction of Solar Constant
0.5
1
1.5
Fraction of Solar Constant
e) Given that the Sun was fainter in the past, what do these diagrams “predict” the current
Earth state should be?
If the Sun’s intensity has increased monotonically the Earth ought to be stuck in a Snowball
state.
What missing processes might resolve this “faint young Sun paradox”?
The absence of carbon cycle feedbacks may result in the Snowball Earth being spuriously
stable. In reality, the weathering of silicate rocks, which removes CO2 from the atmosphere,
would be “shut-off” by the ice such that volcanic emissions would drive-up the atmospheric
concentration of CO2. The resulting greenhouse effect may have destabilised the Snowball
Earth.
Other possibilities for avoiding the Snowball state include negative feedbacks via changes in
poleward heat transport (e.g. Maximum Entropy Production principle), or biological
regulation of the climate (i.e. Gaian regulation).
4. Return parameters to their default values in table 1.
a) What is the impact of doubling CO2 on global temperature and albedo?
T = 3.45 oC ,  = 0.0092
b) Why does this differ from the T2x input parameter?
Because of positive ice-albedo feedbacks, Tsnow = 3.45 – 2.0 = 1.45 oC
c) The global mean temperature is given by :
<T> = [ <RS> {1 - <>} – A ] / B
where <RS> =1370 fS / 4 is the mean incident solar radiation, <> is the mean albedo, and
B=2.17 W m-2 oC-1. Derive a relation between the increment to the mean temperature due to
albedo changes, Tsnow , and the albedo change itself, .
Tsnow = <RS>  / B = 157.8 
Does this fit with the answers given in (a) and (b) above?
Yes.
d) Now try 4xCO2 in Run 1. How large are the mean temperature and mean albedo changes in
going from 2xCO2 to 4xCO2? Why are they lower than in (a)?
Because at double CO2 no ice is left to melt, so no further albedo feedback.