Question

If the global albedo is increased from 0.3 to 0.4, the global radiative equilibrium temperature (in K) would decrease by .......... (rounded off to one decimal place). (Consider no greenhouse effect, solar constant = 1360 W/m$^2$, Stefan–Boltzmann constant = $5.67 \times 10^{-8}$ W m$^{-2}$ K$^{-4}$).

Hint:

Higher albedo → less solar absorption → cooler equilibrium temperature.

Answer 1

Correct Answer: 9.5

Step 1: Recall equilibrium temperature formula.
\[ T = \left( \frac{S (1-\alpha)}{4 \sigma} \right)^{1/4} \] where, $S = 1360 \, \text{W/m}^2$, $\alpha =$ albedo, $\sigma = 5.67 \times 10^{-8}$.
Step 2: Calculate for $\alpha = 0.3$.
\[ F = \frac{S (1-\alpha)}{4} = \frac{1360 \times 0.7}{4} = \frac{952}{4} = 238 \, \text{W/m}^2 \] \[ T_1 = \left( \frac{238}{5.67 \times 10^{-8}} \right)^{1/4} \] \[ = (4.20 \times 10^9)^{1/4} \approx 254.9 \, K \]
Step 3: Calculate for $\alpha = 0.4$.
\[ F = \frac{1360 \times 0.6}{4} = \frac{816}{4} = 204 \, \text{W/m}^2 \] \[ T_2 = \left( \frac{204}{5.67 \times 10^{-8}} \right)^{1/4} \] \[ = (3.60 \times 10^9)^{1/4} \approx 244.0 \, K \]
Step 4: Find decrease.
\[ \Delta T = T_1 - T_2 = 254.9 - 244.0 = 10.9 \, K \] Final Answer: \[ \boxed{10.9 \, K} \]