A coal-fired power station has a power output of 4.0GW. It has been suggested that a wind farm could replace this power station. Using the data below, determine the area that the wind farm would occupy in order to meet the same power output as the coal-fired
power station.

Radius of wind turbine blades = 42 m
Area required by each turbine = 5.0×10⁴ m²
Efficiency of a turbine = 30%
Average annual wind speed = 12 m s^–1
Average annual density of air = 1.2 kg m^–3

Wind power does not involve the production of greenhouse gases. Outline why the surface temperature of the Earth is higher than would be expected without the greenhouse effect.

The average solar intensity incident at the surface of the Earth is 238 W m^–2.

(i) Assuming that the emissivity of the surface of the Earth is 1.0, estimate the average surface temperature if there were no greenhouse effect.

(ii) The enhanced greenhouse effect suggests that in several decades the predicted temperature of the atmosphere will be 250 K. The emissivity of the atmosphere is 0.78. Show that this atmospheric temperature increase will lead to a predicted average Earth surface temperature of 292 K.

power output of a turbine=0.3×½ρAv³=0.3×0.5×1.2×3.14×[42]²×[12]³(=1723kW);
number of turbines needed \( = \frac{{4 \times {{10}^9}}}{{1.723 \times {{10}^6}}}\left( { = 2322} \right)\);
area needed = 2322×5.0×10⁴;
=1.2×10⁸(m²);
Award [4] for a bald correct answer.
Note: Answers sometimes start with calculating power input from wind which is 5743 kW and incorporate 0.3 at a later stage.

look for these main points:
the surface of Earth re-radiates the Sun’s radiation;
greenhouse gases (in atmosphere) readily absorb infrared;
mention of resonance;
the absorbed radiation is re-emitted (by atmosphere) in all directions;
(some of) which reaches the Earth and further heats the surface;
Award [1 max] for responses along the lines that greenhouse gases trap infrared radiation.

(i) total absorbed radiation= total emitted radiation =238(Wm ^-2);
temperature of Earth=\({\left[ {\frac{{238}}{{5.67 \times {{10}^{ - 8}}}}} \right]^{\frac{1}{4}}}\)=255(K);
Award [2] for a bald correct answer.

(ii) total absorbed radiation at surface=238+[(εδT⁴)0.78×5.67×10^-8×250⁴];
=410.8(Wm^-2);
temperature of surface=\({\left[ {\frac{{410.8}}{{5.67 \times {{10}^{ - 8}}}}} \right]^{\frac{1}{4}}}\)=291.7(K);
≈292K

This was another calculation in which candidates are becoming well versed. There are a number of steps and many were able to negotiate them with ease. Failures included omitting the efficiency or getting it the wrong way up in the equation. Although full marks were given for the correct answer candidates would be well advised in such questions, to fully explain each step in their argument so that part-credit can be obtained. A jumble of arithmetic with the wrong answer will score zero.

A sizeable majority talked about the infrared radiation being trapped in the atmosphere. This did not attract full credit as it fails to grasp the nettle of the interaction between the earth’s surface and the atmosphere. A generous number of points were available on the scheme but most gained two out of three marks. This question was poorly answered by Spanish-speaking candidates.

(i) temperature of surface

(ii)This was another “show that” question. Candidates need to display reasoning - more able candidates could satisfy examiners on this point.