Date | May 2012 | Marks available | 4 | Reference code | 12M.2.SL.TZ1.2 |
Level | Standard level | Paper | Paper 2 | Time zone | Time zone 1 |
Command term | Calculate and Explain | Question number | 2 | Adapted from | N/A |
Question
This question is about the greenhouse effect.
The following data are available for use in this question:
Explain why the power absorbed by the Earth is
\[\frac{P}{{4\pi {d^2}}} \times \left( {1 - \alpha } \right) \times \pi {r^2}\]
The equation in (a) leads to the following expression which can be used to predict the Earth’s average surface temperature T.
\[T = \sqrt[4]{{\frac{{\left( {1 - \alpha } \right)P}}{{16\pi \sigma {d^2}}}}}\]
(i) Calculate the predicted temperature of the Earth.
(ii) Explain why the actual average surface temperature of the Earth is in fact higher than the answer to (b)(i).
Markscheme
intensity of the Sun’s radiation at the Earth’s orbit =\(\frac{P}{{4\pi {d^2}}}\);
fraction absorbed by the Earth =(1-∝);
the surface area of the disc (absorbing the radiation) \( = \pi {r^2}\);
Look for statements that correctly describe each term.
(i) correct substitution;
to get T=250 (K);
(ii) greenhouse gases in the atmosphere absorb some of the energy radiated by the Earth;
and radiate some of it back to the surface of the Earth;