Heat engines & refrigerators: practice problems with step-by-step solutions
Free Heat engines & refrigerators practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: first law of a cycle, Carnot limit, refrigerators and heat pumps.
Free practice problemEasy
A heat pump with a coefficient of performance of 2.9 delivers 17 kW of heat to a house. Find the rate Q̇L at which it takes heat from the outside air.
A heat engine turns heat into work. A car engine, a steam turbine in a power plant and a jet engine are all heat engines.
Every heat engine works the same way. It takes heat QH from something hot, turns part of it into work W, and dumps the rest QC into something cold.
The hot and cold places are called reservoirs: a flame, a boiler, the outside air, a river. They are so large that their temperatures do not change.
An engine runs in a cycle: after each turn it is back in the same state. So energy cannot pile up inside it. What goes in must come out:
QH = W + QC
Self-check An engine takes in 100 kJ of heat per cycle and produces 40 kJ of work. How much heat does it reject?
140 kJ
2.5 kJ
40 kJ
60 kJ
Answer: 60 kJ. Energy in equals energy out: QC = QH − W = 100 − 40 = 60 kJ. Adding them (140 kJ) counts the work as an input.
2Efficiency and the Carnot limit
The thermal efficiency is the share of the heat input that becomes work:
η = W / QH = 1 − QC / QH
No engine can turn all its heat into work. There is always some heat to reject. This is the second law of thermodynamics.
The best possible engine between two temperatures is the ideal Carnot engine. Its efficiency depends only on the reservoir temperatures, in kelvin:
ηmax = 1 − TC / TH
Always convert to kelvin first: T(K) = T(°C) + 273.15. With °C the formula gives nonsense.
A real engine always has η < ηmax. An engine that claims more than the Carnot limit is impossible.
Self-check What is the highest possible efficiency of an engine working between 327 °C and 27 °C?
100%
50%
8.3%
91.7%
Answer: 50%. In kelvin: 600.15 K and 300.15 K, so ηmax = 1 − 300.15/600.15 = 50%. Putting the °C values in the formula gives 1 − 27/327 = 91.7%, which is wrong.
3Refrigerators and heat pumps
A refrigerator runs an engine backwards. Work W goes in, and heat is pumped from the cold side to the hot side.
A heat pump is the same machine, used for heating: it takes heat from the cold outside air and delivers it to a warm house. The energy balance is still QH = W + QC.
Instead of an efficiency, we use the coefficient of performance (COP): what you want, divided by the work you pay for. It is usually larger than 1.
COPR = QC / W COPHP = QH / W = COPR + 1
The Carnot limits, again with temperatures in kelvin:
Self-check A heat pump delivers 6 kW of heat to a house while using 2 kW of electricity. How much heat does it take from the outside air?
12 kW
3 kW
4 kW
8 kW
Answer: 4 kW. QC = QH − W = 6 − 2 = 4 kW. Its COP is 6/2 = 3, so 3 kW is the COP, not a heat flow.
4The method step by step
Sketch the device between its two reservoirs and draw arrows for QH, QC and W. Decide whether it is an engine or a refrigerator / heat pump.
Write the energy balance QH = W + QC. It holds for every device, ideal or not. Rates in kW work the same way as amounts in kJ.
If temperatures are given, convert them to kelvin.
Use the right measure: η = W/QH for an engine, COPR = QC/W for a refrigerator, COPHP = QH/W for a heat pump.
For the best possible device, use the Carnot formula. Then the unknown heat or work follows from the energy balance.
“Most work”, “highest possible efficiency”, “minimum power input” and “reversible” all point to the Carnot limit.
5A worked example
Follow every step. Try to predict the next one before you open it.
Worked example
A heat engine works between a hot reservoir at 500 °C and a cold reservoir at 25 °C. It takes in 3000 kJ of heat per cycle. Find the most work it could possibly produce. An inventor claims 2000 kJ per cycle: is that possible?
1
Draw the engine
The most work comes from an ideal Carnot engine between the same two reservoirs.
2
Convert to kelvin
TH = 500 + 273.15 = 773.15 K
TC = 25 + 273.15 = 298.15 K
3
The Carnot efficiency
ηmax = 1 − 298.15 / 773.15 = 1 − 0.3856 = 0.614
At best, 61.4% of the heat can become work.
4
The most work
Wmax = ηmax · QH = 0.6144 · 3000 = 1843 kJ
5
Check the answer
The ideal engine rejects QC = 3000 · 298.15/773.15 = 1157 kJ. Then W + QC = 1843 + 1157 = 3000 kJ, which matches the heat input.
6
Judge the claim
2000 kJ is more than the 1843 kJ that even a perfect engine could deliver. The claim breaks the second law, so it is impossible.
Answer
Wmax ≈ 1843 kJ per cycle (ηmax ≈ 61.4%), so a claim of 2000 kJ is impossible.
6Common mistakes
Putting temperatures in °C into the Carnot formulas. Always use kelvin.
Mixing up the two COPs. A refrigerator counts QC (heat removed), a heat pump counts QH (heat delivered).
Using the Carnot efficiency for a real engine when the heats are given. Then η = W/QH from the actual numbers.
Forgetting that the work is also delivered as heat by a refrigerator: QH = QC + W, so the kitchen gets more heat than the fridge removes.
Writing an efficiency as 61.4 instead of 0.614 inside a calculation.
Self-check What is the highest possible COP of a refrigerator that keeps its inside at 0 °C in a kitchen at 27 °C?
0.91
0
10.1
11.1
Answer: 10.1. COPR,max = TC/(TH − TC) = 273.15/27 = 10.1. Using TH on top gives the heat-pump value 11.1, and using 0 °C directly gives 0.
7How exam questions look
Easy: the first law of a cycle. From two of QH, W and QC, find the third or the efficiency. Variations use power plants (MW), refrigerators and heat pumps with a given COP.
Medium: the Carnot limit. Find ηmax or Wmax from the reservoir temperatures. Others ask for a second-law efficiency (η/ηmax), the cold reservoir temperature of a reversible engine, or the heat rejected by a reversible refrigerator.
Hard: refrigerators and heat pumps at the Carnot limit: find COPmax or the minimum power input. Others chain two devices, or ask for the entropy generated by a real engine.
For the minimum power input: first COPmax from the temperatures, then Wmin = QH/COPHP,max or QC/COPR,max.
For entropy generated per cycle: Sgen = QC/TC − QH/TH, which must be zero or positive.
A heat pump with a coefficient of performance of 5 delivers 4 kW of heat to a house. Find the rate Q̇L at which it takes heat from the outside air.
COPHP = 5Q̇H = 4 kW
Find Q̇L(kW)
Show hints
A heat pump’s useful output is the heat delivered: COPHP = Q̇H/Ẇ.
Energy balance over the cycle: Q̇L = Q̇H − Ẇ.
Ẇ = 0.8 kW.
Show the answer and the step-by-step solution
Second-law limits: no cycle working between reservoirs at absolute temperatures TH and TL can beat the reversible (Carnot) cycle: ηmax = 1 − TL/TH, COPR,max = TL/(TH − TL) and COPHP,max = TH/(TH − TL). Energy is always conserved: QH = W + QL.
Power input
Ẇ = Q̇H/COP = 4/5 = 0.8 kW
Heat taken from outside
Q̇L = Q̇H − Ẇ = 4 − 0.8 = 3.2 kW
Answer
Q̇L = 3.2 kW
Example 2Medium
A reversible (Carnot) heat engine takes in 3100 kJ of heat from a reservoir at 625 °C and rejects 977 kJ per cycle. Find the temperature TL of the cold reservoir in °C.
TH = 625 °CQH = 3100 kJQL = 977 kJ
Find TL(°C)
Show hints
For a reversible cycle the heats are in the ratio of the absolute temperatures: QL/QH = TL/TH.
Use absolute temperatures: K = °C + 273.15.
TH = 898.1 K.
Show the answer and the step-by-step solution
Second-law limits: no cycle working between reservoirs at absolute temperatures TH and TL can beat the reversible (Carnot) cycle: ηmax = 1 − TL/TH, COPR,max = TL/(TH − TL) and COPHP,max = TH/(TH − TL). Energy is always conserved: QH = W + QL.
Absolute temperature of the source
TH = 625 + 273.15 = 898.1 K
Reversible heat ratio
TL = TH·QL/QH = 898.1·977/3100 = 283.1 K = 9.912 °C
Answer
TL = 9.912 °C
Example 3Hard
A heat pump keeps a house at 20 °C while it is 2 °C outside, delivering 14 kW of heat to the house. Find the minimum power input Ẇmin.
TH = 20 °CTL = 2 °CQ̇H = 14 kW
Find Ẇmin(kW)
Show hints
A heat pump’s COP is the heat delivered divided by the work in: COP = QH/W. The reversible limit is TH/(TH − TL).
Use absolute temperatures: K = °C + 273.15.
COP = 16.29.
Show the answer and the step-by-step solution
Second-law limits: no cycle working between reservoirs at absolute temperatures TH and TL can beat the reversible (Carnot) cycle: ηmax = 1 − TL/TH, COPR,max = TL/(TH − TL) and COPHP,max = TH/(TH − TL). Energy is always conserved: QH = W + QL.
The least work is needed by a reversible machine, which has the highest COP.
Ẇmin = Q̇H/COP = 14/16.29 = 0.8596 kW
Answer
Ẇmin = 0.8596 kW
Questions and answers
What is Heat engines & refrigerators?
A heat engine turns heat into work. A car engine, a steam turbine in a power plant and a jet engine are all heat engines.
Every heat engine works the same way. It takes heat QH from something hot, turns part of it into work W, and dumps the rest QC into something cold.
How do I solve Heat engines & refrigerators problems step by step?
Second-law limits: no cycle working between reservoirs at absolute temperatures TH and TL can beat the reversible (Carnot) cycle: ηmax = 1 − TL/TH, COPR,max = TL/(TH − TL) and COPHP,max = TH/(TH − TL). Energy is always conserved: QH = W + QL.
What are common mistakes in Heat engines & refrigerators?
COP = Q<sub>H</sub>/W gives the work; the heat from outside is then Q<sub>L</sub> = QH − W.
That is the work input. The heat from outside is Q<sub>L</sub> = Q<sub>H</sub> − W.
For a reversible cycle Q<sub>L</sub>/Q<sub>H</sub> = T<sub>L</sub>/T<sub>H</sub> holds only with absolute temperatures. Convert to kelvin first, then back to °C.
What do the Heat engines & refrigerators exercises cover?
Easy: first law of a cycle
Medium: Carnot limit
Hard: refrigerators and heat pumps
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.