1. **Problem 13:** Given the albedo $a=0.25$ and reflected intensity $I_r=240\ \mathrm{Wm^{-2}}$, find the outgoing radiated intensity $I_o$. 2. **Formula and explanation:** Albedo is the fraction of incident radiation reflected by a surface. The reflected intensity is related to the incident intensity $I_i$ by $$I_r = a I_i$$ The outgoing radiated intensity is the part of the incident intensity not reflected, so $$I_o = I_i - I_r = I_i (1 - a)$$ 3. **Calculate incident intensity $I_i$: $$I_i = \frac{I_r}{a} = \frac{240}{0.25} = 960\ \mathrm{Wm^{-2}}$$ 4. **Calculate outgoing radiated intensity $I_o$: $$I_o = I_i (1 - a) = 960 \times (1 - 0.25) = 960 \times 0.75 = 720\ \mathrm{Wm^{-2}}$$ 5. **Answer for problem 13:** The outgoing radiated intensity is $\boxed{720\ \mathrm{Wm^{-2}}}$, which corresponds to option C. --- 6. **Problem 14:** The graph shows internal energy $U$ vs temperature $T$ for an ideal monatomic gas. $U$ increases linearly with $T$. Given $U=4500\ \mathrm{J}$ at $T=27^\circ C$. 7. **Formula and explanation:** For an ideal monatomic gas, internal energy is $$U = \frac{3}{2} nRT$$ where $n$ is moles, $R$ is gas constant, $T$ is temperature in Kelvin. 8. **Convert temperature to Kelvin:** $$T_K = 27 + 273 = 300\ \mathrm{K}$$ 9. **Calculate number of moles $n$: $$n = \frac{2U}{3RT} = \frac{2 \times 4500}{3 \times 8.314 \times 300} = \frac{9000}{7482.6} \approx 1.203$$ 10. **Interpretation:** The sample contains approximately 1.2 moles of gas. --- **Summary:** - Problem 13 answer: $720\ \mathrm{Wm^{-2}}$ (option C). - Problem 14: Internal energy linear with temperature, $n \approx 1.2$ moles.