| Date | May 2012 | Marks available | 3 | Reference code | 12M.2.SL.TZ2.8 |
| Level | Standard level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Show | Question number | 8 | Adapted from | N/A |
Question
This question is in two parts. Part 1 is about solar power and climate models. Part 2 is about gravitational fields and electric fields.
Part 1 Solar power and climate models
Distinguish, in terms of the energy changes involved, between a solar heating panel and a photovoltaic cell.
State an appropriate domestic use for a
(i) solar heating panel.
(ii) photovoltaic cell.
The radiant power of the Sun is 3.90 ×1026W. The average radius of the Earth’s orbit about the Sun is 1.50 ×1011 m. The albedo of the atmosphere is 0.300 and it may be assumed that no energy is absorbed by the atmosphere.
Show that the intensity incident on a solar heating panel at the Earth’s surface when the Sun is directly overhead is 966 Wm–2.
Show, using your answer to (c), that the average intensity incident on the Earth’s surface is 242 Wm–2.
Assuming that the Earth’s surface behaves as a black-body and that no energy is absorbed by the atmosphere, use your answer to (d) to show that the average temperature of the Earth’s surface is predicted to be 256 K.
Markscheme
a solar heating panel converts the (radiation) energy of the Sun into thermal/heat energy;
(allow “solar energy” but do not allow “heat”)
a photovoltaic cell converts the (radiation) energy of the Sun into electrical energy;
(i) water heater / any specific use such as swimming pool/bath;
(ii) powering TV/radio/lighting/any low energy electrical appliance;
surface area of sphere at 1.5×1011m from Sun =4π×1.502×1022;
power per m2=\(\frac{{3.90 \times {{10}^{26}}}}{{4 \times 3.14 \times {{1.50}^2} \times {{10}^{22}}}} = 1.38 \times {10^3}\);
(presence of the substitution allows inference of first marking point)
power per m2 at surface = 0.7×1.38×103Wm-2;
(=966Wm-2)
Earth appears, to the Sun, like a disc of radius R; (must be explicit)
intensity=power incident per unit area; (must be explicit in words or equation)
(power incident per unit area)=\(\frac{{966\pi {R^2}}}{{4\pi {R^2}}}\);
(=242Wm-2)
(power absorbed) 242 = (power emitted) σT4 ;
\(T = {\left[ {\frac{{242}}{{5.67 \times {{10}^{ - 8}}}}} \right]^{\frac{1}{4}}}\) or 255.5;
(=256K)
Examiners report
In an easy opener, candidates were asked for the energy changes in solar heating panels and photovoltaic cells. Sometimes the word “energy” did not appear in the answer. It is important for candidates to give a clear statement of the initial and the final energy forms expressed in scientific language.
(i) and (ii) There was a wide variety of correct responses here. However some are clearly confused about the uses of solar heating panels and photovoltaics.
This was done well with many correctly showing the intensity arriving from the Sun and incorporating the effect of albedo appropriately.
This was not so impressive (as (c)) with many inclusions of the factor of 4 with no explanation of its origin. This was not acceptable.
This was straightforward but a clear manipulation of Stefan‟s Law was required, ideally with a calculation with significant figures quoted to better than the quoted answer. Many failed in this respect by giving an initial substitution and nothing else. Examiners needed to see correct handling of the fourth root to award full credit.
