Why is so little nuclear fuel needed?

A kilogram of uranium yields about 8 times ten to the thirteenth joules, millions of times more than coal.

What does 40 per cent efficiency mean?

Only forty per cent of the thermal power becomes electricity; the rest is dumped through cooling towers.

S6 Physics Nuclear Energy, Paper 2: mass defect and fuel use

2026-11-02

Tutor Prep Guide · Primary English

Ten graded items turning mass defect into energy, then thermal power, daily fuel use and yearly uranium demand.

Focus　Convert every mass to kilograms and every energy to joules before comparing, and divide total energy by energy per kilogram for fuel mass.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	25
Difficulty mix	Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter	Chapter: S6 Physics: Nuclear Energy
Level	Four basic, three advanced, two challenge and one error-repair item

02　Questions (10 items)

FoundationHow much energy comes from converting 0.5 kg of mass completely, with c as 3.0 x 10^8 m/s?
FoundationFind the energy released if 2 kg of matter is converted completely into energy.
FoundationA reaction has a mass defect of 2 u. Find the energy released, with 1 u as 931 MeV.
FoundationA mass defect of 1 g releases how much energy?
IntermediateA reactor runs at 2.0 x 10^9 W thermal power and 40 per cent efficiency. Find the electrical output.
IntermediateEach fission gives 3.2 x 10^-11 J. How many fissions per second keep a 100 W lamp running?
Intermediate1 kg of uranium-235 holds about 2.56 x 10^24 nuclei and each fission gives 3.2 x 10^-11 J. Find the total energy.
ChallengeA plant delivers 1.2 GW of electricity at 40 per cent efficiency. Find the thermal power and the daily uranium use at 8.2 x 10^13 J per kg.
ChallengeA region uses 3.15 x 10^17 J a year. How much uranium is needed at 8.2 x 10^13 J per kg?
Error redoA student gets 9.0 x 10^16 J for a 1 g mass defect. What went wrong and what is the correct value?

03　Answers and one-line notes

#	Answer	One-line note
1	4.5 x 10^16 J	Half a kilogram times the square of the light speed gives 4.5 times ten to the sixteenth joules.
2	1.8 x 10^17 J	Two kilograms double the one kilogram result, giving 1.8 times ten to the seventeenth joules.
3	1862 MeV	Two atomic mass units at 931 MeV each give 1862 MeV.
4	9.0 x 10^13 J	One gram is 0.001 kg, so the energy is 9.0 times ten to the thirteenth joules, still enormous.
5	8.0 x 10^8 W	Forty per cent of two gigawatts is 0.8 GW of electrical output, the rest leaves as waste heat.
6	3.125 x 10^12 per second	One hundred joules per second divided by the energy per fission gives about 3.1 times ten to the twelfth fissions.
7	About 8.2 x 10^13 J	Multiplying the nuclei by the energy per fission gives about 8.2 times ten to the thirteenth joules per kilogram.
8	3.0 x 10^9 W and about 1.3 kg per day	Thermal power is 3.0 GW and the daily output of about 1.04 times ten to the fourteenth joules needs only about 1.3 kg of fuel.
9	About 3.8 x 10^3 kg	Dividing the yearly demand by the energy per kilogram gives about 3.8 tonnes of uranium.
10	He used 1 kg instead of 0.001 kg; the true value is 9.0 x 10^13 J, a thousand times smaller	He substituted 1 for one gram instead of 0.001 kg, so his answer is a thousand times too large.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.