Specification: AQA A-level Physics (7408) 3.11.2.5 Engineering physics option | Paper: 3 Section B | Data and formulae booklet: efficiency = W/QH = (QH − QC)/QH and maximum possible efficiency = (TH − TC)/TH are given in the exam
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Second law of thermodynamics (heat engines): a heat engine must operate between a hot source and a cold sink. It is impossible for an engine working in a cycle to convert all the heat it takes in into work: some heat QC must always be rejected to the cold sink.
Efficiency of a heat engine: the work done by the engine ÷ the heat energy taken in from the hot source, W ÷ QH. By conservation of energy QH = W + QC.
Maximum theoretical efficiency: the efficiency of an ideal (reversible) engine, (TH − TC) ÷ TH, with both temperatures in kelvin. Real engines are less efficient because of friction, heat losses and processes that are not reversible.
The equation
efficiency = work done ÷ heat input efficiency = W ÷ QH = (QH − QC) ÷ QH and maximum possible efficiency = (TH − TC) ÷ TH
| Quantity | Symbol | Unit |
|---|---|---|
| work done by the engine | W | joules (J) |
| heat taken in from the hot source | QH | joules (J) |
| heat rejected to the cold sink | QC | joules (J) |
| temperature of the hot source | TH | kelvin (K) |
| temperature of the cold sink | TC | kelvin (K) |
| efficiency | – | no unit (a decimal or a percentage) |
Rearranged: W = efficiency × QH | QH = W ÷ efficiency | QC = QH − W | TC = TH × (1 − maximum efficiency) | The same equations work with powers (J s⁻¹) in place of energies. Temperatures must be in kelvin: add 273 to °C.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) In each cycle a heat engine takes in 2000 J of heat from a hot source and does 700 J of work. Calculate its efficiency.
Show solution
F: efficiency = W ÷ QH
I: efficiency = 700 ÷ 2000
F: efficiency is already the subject
A: 0.35 (35%)
Q2 (Core) A heat engine works between a hot source at 800 K and a cold sink at 300 K. Calculate its maximum possible efficiency.
Show solution
F: maximum efficiency = (TH − TC) ÷ TH
I: maximum efficiency = (800 − 300) ÷ 800 = 500 ÷ 800
F: maximum efficiency is already the subject
A: 0.625 (62.5%)
Q3 (Core) Steam enters a power station turbine at 540 °C and the condenser is at 30 °C. Calculate the maximum possible efficiency, to 2 significant figures.
Show solution
Convert first: TH = 540 + 273 = 813 K; TC = 30 + 273 = 303 K
F: maximum efficiency = (TH − TC) ÷ TH
I: maximum efficiency = (813 − 303) ÷ 813 = 510 ÷ 813
F: maximum efficiency is already the subject
A: 0.627… = 63% (2 s.f.)
Watch out: using °C gives (540 − 30) ÷ 540 = 94%, which is far too high. Always convert to kelvin first.
Q4 (Core) An engine with an efficiency of 28% does 4.2 kJ of work in each cycle. Calculate the heat taken in and the heat rejected in each cycle.
Show solution
Convert first: 4.2 kJ = 4200 J; 28% = 0.28
F: efficiency = W ÷ QH
I: 0.28 = 4200 ÷ QH
F: QH = 4200 ÷ 0.28 = 15 000 J
A: heat taken in = 15 kJ; heat rejected QC = QH − W = 15 000 − 4200 = 10 800 J = 10.8 kJ
Q5 (Exam) A power station’s turbines produce 500 MW of mechanical power with an efficiency of 38%. The steam enters at 840 K and the condenser is at 300 K. Calculate, to 2 significant figures, (a) the rate of heat input, (b) the rate at which heat is rejected, (c) the maximum possible efficiency, and comment.
Show solution
Convert first: 500 MW = 5.00 × 108 W
Step 1 – (a) heat input per second
QH per second = W per second ÷ efficiency = 5.00 × 108 ÷ 0.38 = 1.315… × 109 W
A: 1.3 × 109 W (2 s.f.)
Step 2 – (b) heat rejected per second
QC = QH − W = 1.315… × 109 − 5.00 × 108 = 8.157… × 108 W
A: 8.2 × 108 W (2 s.f.)
Step 3 – (c) maximum efficiency
(840 − 300) ÷ 840 = 0.642…, so 64% (2 s.f.). The real efficiency is well below this because of friction, heat losses and irreversible processes.
Q6 (Exam) An inventor claims that an engine working between 650 K and 300 K has an efficiency of 60%. Show whether the claim is possible.
Show solution
F: maximum efficiency = (TH − TC) ÷ TH
I: maximum efficiency = (650 − 300) ÷ 650 = 350 ÷ 650
F: maximum efficiency is already the subject
A: 0.538… = 54% (2 s.f.). This is less than 60%, so the claim is impossible: no engine between these temperatures can be more efficient than the theoretical maximum.
Q7 (Exam) A car engine produces a useful mechanical power of 45 kW with an efficiency of 30%. Calculate the energy rejected as heat in one minute, to 2 significant figures.
Show solution
Convert first: 45 kW = 45 000 W; 30% = 0.30
Step 1 – rate of heat input
QH per second = 45 000 ÷ 0.30 = 150 000 W
Step 2 – rate of heat rejection
QC per second = 150 000 − 45 000 = 105 000 W
Step 3 – energy in one minute
105 000 × 60 = 6 300 000 J
A: 6.3 × 106 J (2 s.f.)
Watch out: the heat rejected is QH − W, not W ÷ efficiency. Find QH first, then subtract.
Q8 (Exam) An engine works between 600 K and 300 K. Which raises its maximum possible efficiency more: increasing TH by 50 K or decreasing TC by 50 K? Give each new value to 3 significant figures.
Show solution
Step 1 – original
(600 − 300) ÷ 600 = 0.500, so 50.0%
Step 2 – raise TH to 650 K
(650 − 300) ÷ 650 = 350 ÷ 650 = 0.538…, so 53.8%
Step 3 – lower TC to 250 K
(600 − 250) ÷ 600 = 350 ÷ 600 = 0.583…, so 58.3%
A: lowering TC by 50 K has the bigger effect, because the same temperature difference is divided by the smaller TH.
Q9 (Stretch) An engine completes 25 cycles per second, taking in 1.6 kJ of heat per cycle. It works between 900 K and 320 K, and its actual efficiency is 60% of the maximum possible efficiency. Calculate the output power and the rate of heat rejection, to 3 significant figures.
Show solution
Step 1 – maximum efficiency
(900 − 320) ÷ 900 = 580 ÷ 900 = 0.6444…
Step 2 – actual efficiency
0.60 × 0.6444… = 0.3866…
Step 3 – work per cycle
W = efficiency × QH = 0.3866… × 1600 = 618.6… J
Step 4 – power
P = 618.6… × 25 = 15 466.6… W
A: output power = 15.5 kW (3 s.f.)
Step 5 – heat rejected
QC per cycle = 1600 − 618.6… = 981.3… J; per second = 981.3… × 25 = 24 533.3… W
A: 24.5 kW (3 s.f.)
Q10 (Stretch) A power station has a mechanical output of 1.2 GW at an efficiency of 36%. Steam enters the turbines at 570 °C and the cooling water is at 25 °C. (a) Calculate the maximum possible efficiency, to 2 significant figures. (b) Calculate the rate at which heat is rejected, to 3 significant figures. (c) The rejected heat goes into river water, whose temperature must rise by no more than 8.0 K. Calculate the minimum mass of water needed per second, to 2 significant figures. (Specific heat capacity of water = 4180 J kg⁻¹ K⁻¹)
Show solution
Convert first: TH = 570 + 273 = 843 K; TC = 25 + 273 = 298 K; 1.2 GW = 1.2 × 109 W
Step 1 – (a) maximum efficiency
(843 − 298) ÷ 843 = 545 ÷ 843 = 0.646…, so 65% (2 s.f.)
Step 2 – (b) heat rejected per second
QH per second = 1.2 × 109 ÷ 0.36 = 3.333… × 109 W
QC per second = 3.333… × 109 − 1.2 × 109 = 2.133… × 109 W, so 2.13 × 109 W (3 s.f.)
Step 3 – (c) mass of water per second
Q = mcΔθ, so m per second = 2.133… × 109 ÷ (4180 × 8.0) = 63 795.8… kg s⁻¹
A: 6.4 × 104 kg s⁻¹ (2 s.f.)
Watch out: in (b) use the actual efficiency, not the maximum. The maximum only gives the best a perfect engine could do.
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