Why is nuclear fuel so energy dense?

The factor of c squared is 9.0 x 10^16, so one kilogram of uranium matches thousands of tonnes of coal.

Why divide by efficiency?

Efficiency is output over input, so finding the input from a known output means dividing.

S6 Physics Nuclear Energy, Paper 4: energy comparisons and working backwards

2026-11-02

Ten graded items converting grams and kilograms into energy, then working back to thermal power, fission rate and the mass needed.

Focus　Convert every mass to kilograms first, divide electrical output by efficiency for thermal power, and rearrange E = mc squared for 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 1 g of mass completely, with c as 3.0 x 10^8 m/s?
FoundationA reaction loses 0.25 u of mass. Find the energy released, with 1 u as 931 MeV.
FoundationWhat is the total energy if 5 kg of mass is converted entirely?
FoundationA plant gives 500 MW of electricity at 40 per cent efficiency. Find its thermal power.
IntermediateA reactor runs at 3.0 x 10^9 W of thermal power and each fission gives 3.2 x 10^-11 J. Find the fission rate.
IntermediateFind the energy from complete fission of 1 kg of uranium-235, using 6.02 x 10^23 per mole and 3.2 x 10^-11 J per fission.
IntermediateA plant delivers 600 MW at 30 per cent efficiency. Find the daily uranium use at 8.2 x 10^13 J per kg.
ChallengeA reaction loses 0.3 u. Give the energy in MeV and in joules, with 1 MeV as 1.6 x 10^-13 J.
ChallengeA reaction releases 1.8 x 10^14 J. What mass was converted completely?
Error redoA student uses E = half m c squared for 1 kg and gets 4.5 x 10^16 J. Give the correct value and the factor he is wrong by.

03　Answers and one-line notes

#	Answer	One-line note
1	9.0 x 10^13 J	One gram is 1 x 10^-3 kg, so it yields 9.0 x 10^13 J.
2	232.75 MeV	One quarter of 931 MeV gives 232.75 MeV.
3	4.5 x 10^17 J	Each kilogram gives 9.0 x 10^16 J, so five kilograms give 4.5 x 10^17 J.
4	1.25 x 10^9 W	Divide 5.0 x 10^8 W by 0.4 to get 1.25 x 10^9 W of thermal power.
5	About 9.375 x 10^19 per second	3.0 x 10^9 W divided by 3.2 x 10^-11 J gives about 9.4 x 10^19 fissions each second.
6	About 8.2 x 10^13 J	One kilogram holds about 2.56 x 10^24 nuclei, which multiply to 8.2 x 10^13 J.
7	About 0.63 kg	Daily output of 5.2 x 10^13 J divided by 8.2 x 10^13 J per kilogram needs about 0.63 kg.
8	279.3 MeV, that is 4.47 x 10^-11 J	0.3 u gives 279.3 MeV, which equals 4.47 x 10^-11 J after multiplying by 1.6 x 10^-13.
9	2.0 x 10^-3 kg, that is 2 g	Rearranging E = mc² gives 1.8 x 10^14 divided by 9.0 x 10^16, or 2.0 x 10^-3 kg.
10	9.0 x 10^16 J, so he is 2 times too small	The mass-energy relation has no factor of one half, so the true value is 9.0 x 10^16 J, twice his answer.

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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