Answer 1(a)

With clear skies, \(I_{\text{out}} = 299.~\text{W}/\text{m}^2\). With altostratus clouds, \(I_{\text{out}} = 269.~\text{W}/\text{m}^2\), which is \(30.~\text{W}/\text{m}^2\) less, so altostratus clouds have a warming effect.

With stratus clouds, which are closer to the ground, \(I_{\text{out}} = 288.~\text{W}/\text{m}^2\), which is \(10.~\text{W}/\text{m}^2\) less, so stratus clouds have a warming effect.

Altostratus clouds have a bigger effect on \(I_{\text{out}}\). This is because altostratus clouds are higher in the atmosphere (the tops are at 3 km, compared to 1 km for stratus), so the tops of altostratus clouds are cooler than the tops of stratus clouds, and thus emits less longwave radiation.

2. Starting set the altitude to zero and select “Looking up.” When you are looking up, the model reports the longwave radiation coming down to the surface from the atmosphere and hitting the earth’s surface. This heat is in addition to whatever heat the earth gets from shortwave solar radiation.

   For simplicity, think of this as the conditions at night, when the sun is not shining: without sunlight, the temperature of the ground will be determined by balancing the outgoing heat with the heat radiated downward by the warm atmosphere and clouds.

   • First, note the downward longwave heat flux (\(I_{\text{down}}\)) with clear sky (no clouds or rain).
   • Then turn on altostratus clouds and note the change in \(I_{\text{down}}\).
   • Next, turn on stratus clouds and note the change in \(I_{\text{down}}\).

   Answer the following questions:

   • How does \(I_{\text{down}}\) change when you add clouds?
   • Do the clouds have a heating or cooling effect? Why?
   • Which clouds have a greater heating or cooling effect? Why?

Answer 1(b)

With clear skies, \(I_{\text{down}} = 369.~\text{W}/\text{m}^2\). With altostratus clouds, \(I_{\text{down}} = 432.~\text{W}/\text{m}^2\), which is \(62.~\text{W}/\text{m}^2\) greater, so altostratus clouds have a warming effect.

With stratus clouds, which are closer to the ground, \(I_{\text{down}} = 450.~\text{W}/\text{m}^2\), which is \(81.~\text{W}/\text{m}^2\) greater, so stratus clouds have a warming effect.

Stratus clouds have a bigger effect on \(I_{\text{down}}\). This is because altostratus clouds are higher in the atmosphere (the bottoms are at 2.4 km, compared to 0.33 km for stratus), so the bottoms of altostratus clouds are cooler than the bottoms of stratus clouds, and thus emits less longwave radiation.

Answer for 2(a)

At the bottom, the intensity of longwave radiation is 280.9 W/m² and the intensity of shortwave radiation is 266.3 W/m².

At the top of the atmosphere, the intensity of longwave radiation is 244.8 W/m² and the intensity of shortwave radiation is 95.2 W/m².

2. Next, add 100% high clouds (set “High cloud (fraction)” to 1.0). Record the total gain or loss of of heat, and the amount of shortwave and longwave radiation absorbed by the surface and emitted to space. Clouds affect both longwave and shortwave heat fluxes. Which kind of radiation changed more?

Answer for 2(b)

The biggest change was in outgoing longwave radiation from the top of the atmosphere, which dropped by 114.6 W/m².

3. Now do the same thing for low clouds: set “High cloud (fraction)” to zero and “Low cloud (fraction)” to 1.0. Which kind of radiation changed more, compared to the no-cloud condition?

Answer for 2(c)

The biggest change was in the longwave coming down to the ground, which increased by 55.1 W/m².

4. With the low cloud fraction still set to 1.0 (100%), change the drop radius from 10 to 8 \(\mu\)m. How does this change the heat flux?

Answer for 2(d)

There is very little change in the longwave, but big reduction in shortwave reaching the ground and a big increase in shortwave reflected back to space.

In terms of total heat fluxes, with the 10 \(\mu\)m drops, we had \(I_{\text{in}} = 218.8\) and \(I_{\text{out}} = 238.4\), so the earth loses \(19.6~\mathrm{W}/\mathrm{m}^2\).

For the 8 \(\mu\)m drops, we had \(I_{\text{in}} = 216.7\) and \(I_{\text{out}} = 238.3\), so the earth loses \(21.6~\mathrm{W}/\mathrm{m}^2\).

The difference between large and small drops is \(2.0~\mathrm{W}/\mathrm{m}^2\)

5. Now set the cloud fraction to zero and double the CO₂. How does this change the heat flux? How does the effect of doubling CO₂ compare to the effect of changing the droplet size for the low clouds? You can see how important it is to get the cloud droplet size right in climate models!

Answer for 2(e)

In terms of total heat fluxes, with 400 ppm CO₂, we had \(I_{\text{in}} = 244.8\) and \(I_{\text{out}} = 244.8\), so the earth gains as much heat as it loses.

For the 800 ppm CO₂, we had \(I_{\text{in}} = 244.9\) and \(I_{\text{out}} = 240.7\), so the earth gains \(4.2~\mathrm{W}/\mathrm{m}^2\).

Doubling CO₂ changes the heat balance by \(4.2~\mathrm{W}/\mathrm{m}^2\).

Changing the size of cloud drops affects the heat balance about half as much as doubling CO₂, but CO₂ mostly affects \(I_{\text{out}}\) and droplet size mostly affect \(I_{\text{in}}\)

Answer for 3(a)

When I run the RRTM model with the default parameters, the ground temperature is 284.42 K.

When I double CO₂, initially, this reduces \(I_{\text{out}}\) so the Earth gains energy at a rate of 4.2 W/m². I adjust the surface temperature by trial and error, raising it when the Earth gains energy and reducing it when Earth loses energy, until I find that at 286.9 K, the Earth loses as much energy as it gains.

The earth warmed up by 2.48 K when we doubled CO₂, so this is the climate sensitivity.

2. Set CO₂ back to the default value (400 ppm) and set relative humidity to zero. This turns off the water vapor feedback. Adjust the surface temperature offset until the earth loses as much energy as it gains. Write down this temperature.

   Now double the CO₂ and adjust the surface temperature to bring the heat back into balance.

   • How much did the temperature change?

Answer for 3(b)

When I return RRTM to its default conditions and turn off the water vapor feedback by setting relative humidity to zero, this increases \(I_{\text{out}}\) so the Earth loses energy at a rate of 91.1 W/m². I adjust the surface temperature by trial and error until I find that at 261.55 K, the Earth loses as much energy as it gains.

Then I double CO₂ to 800 ppm and the Earth gains energy at a rate of 3.8 W/m². Again, I adjust the surface temperature by trial and error until I find that at 262.65, the Earth loses as much energy as it gains.

The climate sensitivity for 0% relative humidity is 1.1 K.

3. Compare the climate sensitivity with zero relative humidity to the sensitivity with the default value of 80% relative humidity. The difference is the effect of water vapor feedback.

   • What was the amplification factor of the water vapor feedback (the ratio of the climate sensitivity \(\Delta T_{2\times\COO}\) with water vapor feedback to \(\Delta T_{2\times\COO}\) without it)?

Answer for 3(c)

With relative humidity set to zero (no water-vapor feedback), the climate sensitivity is 1.1. With relative humidity set to 80% (normal water-vapor feedback) the climate sensitivity is 2.48. The amplification factor for water-vapor feedback is the ratio of these numbers

\[\text{amplification factor} = \frac{2.48 K}{1.1 K} = 2.2\]

Water vapor feedback more than doubles the climate sensitivity!