Topic Menu

Cyber-chess a chess instruction site

GCSE Questions: Energy Sources

Q27.

(a) During one year, 1.25 × 1018 J of energy was transferred from the National Grid.

Given that the number of seconds in 1 year = 3.16 × 107 calculate the mean power transfer from the National Grid.

Give your answer to 3 significant figures.

Power = energy/time

1.25 × 1018 J/3.16 × 107

Power = 3.96 x 1010 J/s

P = 3.96 x 1010 W

P = 39.6 GW

[4 marks]

(b) The diagram shows a house with a solar power system.

The solar cells generate electricity.

When the electricity generated by the solar cells is not needed, the energy is stored in a large battery.

(i) The charge flow through the cable between the solar cells and the battery in 24 hours box was 27 000 coulombs. Calculate the mean current in the cable.

t = 24 h = 24 x 602 s

I = Q/t

I = 27,000/(24 x 602)

I = 3.125 x 10-4 A

I = 3.1 x 10-4 A

All of the values are given to 2sf - so the final answer should be given to 2 sf too.

[4 marks]

(ii) At one time, the total power input to the solar cells was 7.8 kW.

The efficiency of the solar cells was 15%

Calculate the useful power output of the solar cells.

Useful power output = 15% of total power input

Useful power output = 0.15 x total power input

Useful power output = 0.15 x 7.8 kW

Useful power output = 1.17 kW

[3 marks]

(c) It is unlikely that all of the electricity that the UK needs can be generated by solar power systems. Explain why.

A really large area of land would need to be covered with solar cells

due to (any one of the following reasons):

the low useful power output/efficiency of the solar cells

number of hours of daylight in UK is too low

low solar intensity (in UK)

material for construction of solar cells and/or lithium batteries is in limited supply

[2 marks]

(Total 13 marks)