USAPhO/Thermodynamics

Heat Engines & the Second Law

Efficiency limits, the Carnot cycle, and why perpetual motion machines fail.

Frequent on exams40 min read32 problems

Burn a kilogram of petrol in a car and you get about 44 million joules of heat. The car delivers maybe a quarter of that to the wheels. The rest leaves as hot exhaust and hot radiator water. That looks like bad engineering, and for a century people tried to fix it. Nobody has. Not because the metallurgy is hard, but because nature refuses.

This lesson is about that refusal: how much work you can extract from heat, where the limit comes from, and how to test any claim against it in about thirty seconds.

What an engine is, in bookkeeping terms

In the previous lesson you tracked a gas through single processes with the first law, $\Delta U = Q - W$ , where $Q$ is heat added to the gas and $W$ is work done by the gas. An engine is just that, run in a loop.

A heat engine takes in heat $Q_H$ from something hot, does net work $W$ , and dumps heat $Q_C$ into something cold. The hot thing is the hot reservoir (a burner, burning fuel, a nuclear core) and "reservoir" means it is big enough that taking heat out doesn't change its temperature. The cold thing is the cold reservoir: the river, the outside air, the sea.

The key structural fact is that an engine runs in a cycle. After one full loop, the working gas is back where it started: same pressure, same volume, same temperature. Internal energy depends only on the state, so over a complete cycle $\Delta U = 0$ . The first law then collapses to something you can use without thinking:

W = Q_H - Q_C

Whatever heat goes in and doesn't come out the bottom, comes out the side as work. That single line solves a surprising number of problems. An engine absorbing 2000 J and exhausting 1500 J does 500 J of net work per cycle, and that is the whole calculation.

Figure 1. The pipes are drawn with widths in proportion to the energy flowing through them. Drag the hot reservoir up and watch the work pipe fatten while the waste pipe thins. Notice it never closes.

Efficiency: what you want over what you paid for

You paid for $Q_H$ : that's the fuel. You wanted $W$ . So the thermal efficiency is

e = \frac{W}{Q_H} = \frac{Q_H - Q_C}{Q_H} = 1 - \frac{Q_C}{Q_H}

It is a pure number between 0 and 1, usually quoted as a percentage. An engine taking in 1000 J and delivering 300 J of work is 30% efficient. Two-thirds of the fuel went out the exhaust pipe.

Read the last form again: $e = 1 - Q_C/Q_H$ . To get $e = 1$ you would need $Q_C = 0$ , an engine that dumps nothing. Every real engine dumps. The question the rest of this lesson answers is how much it must dump.

Worth remembering

Three quantities, one equation between them. Given any two of $Q_H$ , $Q_C$ , $W$ you have the third, and then $e$ for free. Before reaching for anything clever, write down $W = Q_H - Q_C$ and see whether the problem is already finished.

Quick check

An engine absorbs 800 J per cycle and does 250 J of work. How much heat does it reject?

Pick one option first.

The ceiling nobody gets over

Here is the argument Sadi Carnot made in 1824, before anyone knew what heat was.

Suppose you had an engine that beat every other engine running between the same two reservoirs. Run your super-engine forwards to produce work, and use that work to drive an ordinary engine backwards: running an engine backwards is legitimate, it just pushes heat from cold to hot and consumes work, which is what a fridge does. Line the two up so that all the work from the first goes into the second.

Now look at the pair from outside. The work cancels. The two machines together take in no work and produce none. But because the first engine is more efficient, it needed less heat from the hot reservoir than the second one puts back. The net effect of the combination is heat moving from cold to hot, by itself, with nothing else changing.

That never happens. Put a cold spoon in hot soup and the spoon warms; it does not chill further while the soup boils. This observation (heat does not flow from cold to hot on its own) is the second law of thermodynamics, and the argument above shows it forbids the super-engine. No engine between two reservoirs can beat a reversible one, and every reversible engine between the same two reservoirs must have the same efficiency, since otherwise you could run this trick between two of them.

A reversible engine is one you could run backwards through exactly the same states. That needs no friction, no turbulence, and heat transferred only across vanishingly small temperature differences, which means running infinitely slowly. It's an idealisation. It is also a hard ceiling. For such an engine the heats turn out to be in the same ratio as the absolute temperatures:

\frac{Q_C}{Q_H} = \frac{T_C}{T_H}

Feed that into $e = 1 - Q_C/Q_H$ and you get the Carnot efficiency:

e_{\text{max}} = 1 - \frac{T_C}{T_H}

Temperatures in kelvin, always. This is the most useful single equation in thermodynamics, because it needs nothing about the engine, not the fuel, not the gas, not the mechanism. Only the two temperatures.

It also says something bleak. A power plant with a 900 K boiler dumping into 300 K cooling water can never exceed $1 - 300/900 = 66.7\%$ , and in practice reaches about 40%. To do better you have to raise $T_H$ or lower $T_C$ , and $T_C$ is set by the river.

Where this goes wrong

Celsius in the Carnot formula is the most common wrong answer in this topic. An engine between 500 °C and 20 °C is not $1 - 20/500 = 96\%$ efficient. Convert first: 773 K and 293 K, giving $1 - 293/773 = 62\%$ . If an efficiency you compute comes out suspiciously close to 1, check the units before you check anything else.

Worked example 1: testing a claim

An inventor advertises an engine that is 45.0% efficient between reservoirs at 400 K and 250 K. By how many percentage points does the claim exceed the legal maximum?

The maximum is Carnot's:

e_{\text{max}} = 1 - \frac{250}{400} = 1 - 0.625 = 0.375

That is 37.5%. The claim exceeds it by $45.0 - 37.5 = 7.5$ percentage points. Note the phrasing: percentage points, a subtraction of two percentages, not a ratio. If instead you were asked what fraction of the ideal the engine reaches, that would be $45.0/37.5$ , a different number entirely.

Quick check

A Carnot engine rejects heat at 300 K and is 25 percent efficient. What is its hot reservoir temperature?

Pick one option first.

Real cycles, and the area on a PV diagram

Carnot's cycle is two isotherms and two adiabats, and no real engine uses it: it would run too slowly to produce useful power. Real engines use cycles that fit into metal.

Whatever the cycle, the net work per cycle is the area enclosed by the loop on a PV diagram, positive if the loop runs clockwise. This follows from $W = \int P\,dV$ : on the upper branch the gas expands at high pressure and does a lot of work; on the lower branch you compress it back at low pressure, which costs less. The difference is the area between them. A rectangular loop between $1.00\times10^5$ Pa and $3.00\times10^5$ Pa and between $1.00\times10^{-3}$ m³ and $4.00\times10^{-3}$ m³ encloses $(2.00\times10^5)(3.00\times10^{-3}) = 600$ J, and that is the net work, no further physics needed.

The Otto cycle is the petrol engine. Compress the fuel–air mixture adiabatically by a factor $r$ (the compression ratio), burn it at constant volume, let it push the piston back adiabatically, then exhaust at constant volume. Because the two adiabats change the temperature by the same factor $r^{\gamma-1}$ , the heats work out to give

e_{\text{Otto}} = 1 - \frac{1}{r^{\gamma - 1}}

where $\gamma$ is the ratio of heat capacities from the previous lesson: 1.40 for air, 5/3 for a monatomic gas like argon. Higher compression means higher efficiency, which is why compression ratios keep creeping up, and why engines knock when they creep too far.

Figure 2. Raise the compression ratio and the loop stretches left, gaining area. Then switch gamma from 1.40 to 1.67 and watch the efficiency jump without changing the geometry at all.

Two cousins come up often and are worth recognising by name rather than deriving. The Diesel cycle burns at constant pressure instead of constant volume, and its efficiency involves a cutoff ratio as well as a compression ratio. The Brayton cycle is the gas turbine, with $e = 1 - r_p^{(1-\gamma)/\gamma}$ for pressure ratio $r_p$ . If a problem hands you one of these formulas, it wants careful substitution and nothing more.

Worked example 2: Otto against Carnot

An Otto cycle with $r = 10.0$ and $\gamma = 1.40$ runs between a lowest temperature of 300 K and a highest of 2000 K. What percentage of the Carnot efficiency does it achieve?

Two efficiencies, then a ratio. Start with the Otto cycle:

e_{\text{Otto}} = 1 - \frac{1}{10.0^{0.40}}

Now $10^{0.40} = 10^{2/5}$ , which is 2.5119. So $1/2.5119 = 0.3981$ , and $e_{\text{Otto}} = 0.6019$ .

The Carnot engine between the same extreme temperatures gets

e_{\text{Carnot}} = 1 - \frac{300}{2000} = 1 - 0.15 = 0.85

The ratio is $0.6019/0.85 = 0.7081$ , so the Otto cycle reaches 70.81% of the Carnot efficiency. The lesson buried in that number: the Otto cycle loses a third of what is theoretically available, not because of friction (this is the ideal Otto cycle) but because it takes in heat over a wide range of temperatures instead of all at $T_H$ .

The Stirling cycle does better. It expands isothermally at $T_H$ , cools at constant volume, compresses isothermally at $T_C$ , and reheats at constant volume. The two isothermal legs each give work $nRT\ln(V_f/V_i)$ , straight from the previous lesson, and the constant-volume legs do no work at all. So for an expansion from $V_0$ to $2V_0$ ,

W_{\text{net}} = nRT_H\ln 2 - nRT_C\ln 2 = nR(T_H - T_C)\ln 2

With 1.00 mol, 600 K and 300 K: $(1.00)(8.31)(300)(0.6931) = 1728$ J per cycle. If a regenerator stores the heat released during the cooling leg and gives it back during the reheat, none of that heat has to be paid for, and the Stirling cycle hits the Carnot efficiency exactly.

Figure 3. Two isotherms closed off by two vertical constant-volume legs. Drag the hot isotherm and watch the net work in the readout grow in proportion to the temperature difference, not to the temperature itself.

Running the engine backwards

Reverse every arrow. Put work in, and heat moves from cold to hot. That is a refrigerator or a heat pump: the same machine, judged by different outputs.

For a refrigerator you want the heat pulled out of the cold box, so the coefficient of performance is

\text{COP}_{\text{fridge}} = \frac{Q_C}{W}

For a heat pump you want the heat delivered into the house, so $\text{COP}_{\text{pump}} = Q_H/W$ . Both are ratios of what you want to what you paid, like efficiency, but both are routinely bigger than 1: a fridge removing 400 J for 100 J of work has COP 4. Nothing is violated; you're moving heat, not creating it.

Since $Q_H = Q_C + W$ , dividing by $W$ gives a relation worth memorising: $\text{COP}_{\text{pump}} = \text{COP}_{\text{fridge}} + 1$ . A machine rated at 4.50 as a fridge is 5.50 as a heat pump.

The Carnot ceiling turns into a floor on the work required. Using $Q_C/Q_H = T_C/T_H$ again,

\text{COP}_{\text{fridge, max}} = \frac{T_C}{T_H - T_C}

A fridge holding 270 K in a 300 K kitchen can at best have COP $270/30 = 9.0$ . Notice what happens as the gap shrinks: the COP blows up, which is why heat pumps are cheap to run in mild weather and expensive in a hard frost.

Where this goes wrong

In the fridge COP the denominator is the temperature difference, not the hot temperature. Writing $T_C/T_H$ gives 0.9 instead of 9.0 in the example above: an answer ten times too small, and less than 1, which should immediately look wrong for a fridge.

Worked example 3: an engine driving a fridge

A Carnot engine absorbs 1000 J from an 800 K reservoir and rejects at 300 K. All its work drives a Carnot refrigerator pumping heat from a 250 K cold box into the same 300 K surroundings. How much heat leaves the cold box?

Nothing here is new; it is two machines chained, and the link between them is the work. Do them one at a time.

The engine first. Its efficiency is

e = 1 - \frac{300}{800} = 1 - 0.375 = 0.625

so the work it produces is $W = (0.625)(1000\ \text{J}) = 625$ J.

Now the fridge, which gets that 625 J. Its best possible COP between 250 K and 300 K is

\text{COP} = \frac{250}{300 - 250} = \frac{250}{50} = 5.00

and COP is $Q_C/W$ , so

Q_C = (5.00)(625\ \text{J}) = 3125\ \text{J}

Over 3000 J out of the cold box for 1000 J of fuel. That is not magic: the 3125 J plus the 625 J of work all end up in the 300 K surroundings, which also received 375 J from the engine's exhaust.

The same chaining logic handles two engines in series. A Carnot engine from 900 K to 600 K, feeding everything it rejects into a second from 600 K to 300 K, is exactly equivalent to one Carnot engine from 900 K to 300 K, efficiency $1 - 300/900 = 66.7\%$ . The intermediate temperature cancels, which is worth remembering: it is the reason the Carnot efficiency can only depend on the two end temperatures.

Worked example 4: when the reservoirs are finite

Two identical blocks of heat capacity $C = 5000$ J/K sit at 400 K and 300 K. A reversible engine runs between them until they reach a common temperature. How much work comes out?

This one takes a real idea. The blocks are not reservoirs: as the engine runs, the hot one cools and the cold one warms, so the efficiency falls to zero by the time they meet. You cannot use a single Carnot efficiency.

What you can use is the property that makes an engine reversible. For a reversible engine the heats satisfy $Q_C/Q_H = T_C/T_H$ , which rearranged is $Q_H/T_H = Q_C/T_C$ . The quantity $Q/T$ taken from the hot side equals the quantity $Q/T$ given to the cold side. That quantity is called entropy, and reversible operation means the total entropy doesn't change. Taking heat $C\,dT$ out of a block at temperature $T$ changes its entropy by $C\,dT/T$ , so summing over the whole run from start to the final temperature $T_f$ :

C\ln\frac{T_f}{400} + C\ln\frac{T_f}{300} = 0

The $C$ divides out, and a sum of logs is the log of a product, so $T_f^2/(400 \times 300) = 1$ . The final temperature is the geometric mean, not the average:

T_f = \sqrt{(400)(300)} = \sqrt{120000} = 346.41\ \text{K}

Now energy. The hot block gives up $C(400 - T_f)$ and the cold block absorbs $C(T_f - 300)$ ; the difference is the work:

W = C(400 - T_f) - C(T_f - 300) = C(700 - 2T_f)

W = 5000(700 - 692.82) = 5000(7.18) = 35900\ \text{J}

Sanity check: the arithmetic mean would have been 350 K, and the geometric mean is lower. The engine has pulled energy out of the pair, so their total must be less than if you had simply let them touch. It is, by exactly 35.9 kJ.

Recognising which tool a problem wants

The problems in this topic sort into four kinds, and the first line of the question usually tells you which.

Only joules mentioned, no temperatures. Use $W = Q_H - Q_C$ and $e = W/Q_H$ . That is the whole toolkit. Power questions live here too: multiply work per cycle by cycles per second, or work backwards from a fuel burn rate times an energy content.

Two temperatures mentioned. Something is being compared to Carnot. Compute $1 - T_C/T_H$ immediately, in kelvin. If the problem also gives a real efficiency, it wants either the gap ("by how many percentage points") or the ratio ("what fraction of the ideal"). Read which.

A compression ratio, a pressure ratio, or a cycle by name. Otto, Diesel, Brayton, Stirling. Use the cycle's own efficiency formula, and expect the problem to then compare it to Carnot between the extreme temperatures.

The words fridge, heat pump, cold box, or a COP. Decide what the machine's user wants (heat removed or heat delivered) before writing anything. Then $\text{COP} = Q/W$ with the right $Q$ on top, and the ceiling $T_C/(T_H - T_C)$ for a fridge.

And one signal that overrides all of these: if the hot or cold object is described with a heat capacity rather than a fixed temperature, it is not a reservoir. Its temperature will drift, and you will have to track $Q/T$ through the whole process as in the last example. That is the bridge to the next lesson, where $Q/T$ gets a name of its own and the second law stops being a rule about engines and becomes a statement about everything.

Problems

Work these on paper, then check your answer here.

Carnot ceiling

NormalWritten for this guidecycles

An engine runs between 600 K and 300 K. What is the maximum possible efficiency, as a percentage?

Efficiency from work and heat input

EasyWritten for this guideefficiency

An engine takes in $1000$ J of heat from its hot reservoir each cycle and delivers $300$ J of work. What is its thermal efficiency, in percent?

Carnot efficiency between two reservoirs

EasyWritten for this guidecarnot

A Carnot engine runs between reservoirs at $500$ K and $300$ K. What is its efficiency, in percent?

Heat dumped to the cold reservoir

EasyWritten for this guidefirst lawengines

Each cycle an engine absorbs $800$ J of heat and performs $250$ J of work, returning to its starting state. How much heat does it reject to the cold reservoir per cycle, in J?

Coefficient of performance of a fridge

EasyWritten for this guiderefrigeratorcop

A refrigerator removes $400$ J of heat from its interior for every $100$ J of electrical work supplied. What is its coefficient of performance (dimensionless)?

dimensionless

Coefficient of performance of a heat pump

EasyWritten for this guideheat pumpcop

A heat pump delivers $600$ J of heat into a house for every $150$ J of work its compressor consumes. What is its coefficient of performance (dimensionless)?

dimensionless

Best possible fridge COP

EasyWritten for this guiderefrigeratorcarnot

A refrigerator keeps its interior at $270$ K while the kitchen is at $300$ K. What is the largest coefficient of performance allowed by thermodynamics (dimensionless)?

dimensionless

Power output from cycles per second

EasyWritten for this guidepowerengines

An engine does $500$ J of net work per cycle and completes $20.0$ cycles each second. What is its power output, in W?

Net work from heat in and out

EasyWritten for this guidefirst lawengines

During one cycle an engine absorbs $2000$ J from a burner and exhausts $1500$ J to the surroundings. How much net work does it do per cycle, in J?

10.

Required hot reservoir temperature

EasyWritten for this guidecarnot

A Carnot engine rejects heat to a reservoir at $300$ K and has an efficiency of $25.0\%$ . What is the temperature of its hot reservoir, in K?

11.

Otto cycle efficiency

EasyWritten for this guideotto cycle

A gasoline engine runs an ideal Otto cycle with compression ratio $8.00$ using air with $\gamma = 1.40$ . What is its efficiency, in percent?

12.

Diesel cycle efficiency

NormalWritten for this guidediesel cycle

An ideal Diesel cycle has compression ratio $r = 18.0$ and cutoff ratio $r_c = 2.00$ , with $\gamma = 1.40$ . Its efficiency is $e = 1 - \frac{1}{\gamma}\frac{r_c^{\gamma}-1}{r^{\gamma-1}(r_c-1)}$ . What is the efficiency, in percent?

13.

Brayton cycle for a gas turbine

NormalWritten for this guidebrayton cycle

A gas turbine runs an ideal Brayton cycle with pressure ratio $10.0$ and $\gamma = 1.40$ . Its efficiency is $e = 1 - r_p^{(1-\gamma)/\gamma}$ . What is the efficiency, in percent?

14.

Testing an inventor's claim★

NormalWritten for this guidesecond lawcarnot

An inventor claims an engine that absorbs $1000$ J from a $600$ K reservoir, rejects heat to a $300$ K reservoir, and produces $600$ J of work per cycle. Working between the same two reservoirs, what is the largest work output per cycle allowed by the second law, in J?

15.

Second-law efficiency of a real engine

NormalWritten for this guideefficiencycarnot

A real engine achieves $30.0\%$ efficiency while a Carnot engine between the same reservoirs would achieve $50.0\%$ . What fraction of the ideal efficiency does the real engine reach, in percent?

16.

Waste heat from a power plant

NormalWritten for this guideefficiencypower

A power plant delivers $1000$ MW of electrical power at $35.0\%$ thermal efficiency. At what rate must it dump waste heat into its cooling water, in MW?

17.

Ideal Stirling engine efficiency

NormalWritten for this guidestirlingcarnot

A Stirling engine with a perfect regenerator runs between a $900$ K burner and $300$ K cooling water. With ideal regeneration it matches the Carnot efficiency. What is that efficiency, in percent?

18.

Power from fuel burn rate

NormalWritten for this guideefficiencypower

A car engine is $25.0\%$ efficient and burns gasoline of energy content $4.00\times10^7$ J/kg at $2.00$ g/s. What is its mechanical power output, in W?

19.

Work needed by a refrigerator

NormalWritten for this guiderefrigeratorcop

A refrigerator with coefficient of performance $3.00$ removes $500$ J of heat from the food compartment each cycle. How much work does the compressor do per cycle, in J?

20.

Relating the two COPs

NormalWritten for this guideheat pumpcop

The same machine can be rated as a refrigerator or as a heat pump. Run in a given environment its refrigerator COP is $4.50$ . What is its heat-pump COP (dimensionless)?

dimensionless

21.

Electrical power drawn by a heat pump

NormalWritten for this guideheat pumppower

A heat pump with coefficient of performance $3.50$ must deliver $12.0$ kW of heat into a house. What electrical power does it consume, in W?

22.

Work per cycle from efficiency

NormalWritten for this guideefficiencyotto cycle

An Otto-cycle engine of efficiency $55.0\%$ absorbs $800$ J of heat during combustion in each cycle. How much net work does it produce per cycle, in J?

23.

Work as the area of a PV rectangle★

NormalWritten for this guidepv diagramwork

An engine runs a rectangular cycle on a $PV$ diagram: pressures alternate between $1.00\times10^5$ Pa and $3.00\times10^5$ Pa, and volumes between $1.00\times10^{-3}$ m $^3$ and $4.00\times10^{-3}$ m $^3$ , traversed clockwise. How much net work does the gas do per cycle, in J?

24.

Carnot engine driving a Carnot fridge★

HardWritten for this guidecarnotrefrigerator

A Carnot engine absorbs $1000$ J from an $800$ K reservoir and rejects heat at $300$ K. All of its work drives a Carnot refrigerator that pumps heat from a $250$ K cold box to the same $300$ K surroundings. How much heat is removed from the cold box, in J?

25.

Two Carnot engines in series

HardWritten for this guidecarnotsecond law

A Carnot engine works between $900$ K and $600$ K, and all the heat it rejects at $600$ K is fed into a second Carnot engine that rejects at $300$ K. Treating the pair as one engine between $900$ K and $300$ K, what is the combined efficiency, in percent?

26.

Otto cycle with a monatomic working gas

HardWritten for this guideotto cyclegamma

An idealized Otto engine with compression ratio $9.00$ uses argon ( $\gamma = 5/3$ ) instead of air as its working gas. What is its efficiency, in percent?

27.

Ocean thermal energy conversion

HardWritten for this guidecarnotpower

An ocean thermal plant uses $300$ K surface water as its hot reservoir and $277$ K deep water as its cold reservoir, and operates at the Carnot limit while producing $1.00$ MW of electric power. At what rate must it reject heat into the deep water, in MW?

28.

By how much does the claim break the second law?

HardWritten for this guidesecond lawcarnot

An inventor advertises an engine that is $45.0\%$ efficient operating between reservoirs at $400$ K and $250$ K. By how many percentage points does the claimed efficiency exceed the maximum permitted by the second law? Give the answer in percentage points.

percentage points

29.

Net work of a Stirling cycle

HardWritten for this guidestirlingwork

A Stirling engine uses $1.00$ mol of an ideal gas. It expands isothermally at $600$ K from $V_0$ to $2V_0$ , cools at constant volume, compresses isothermally at $300$ K back to $V_0$ , and reheats at constant volume. Use $R = 8.31$ J/mol $\cdot$ K. What is the net work per cycle, in J?

30.

Work from two finite bodies★

Very HardWritten for this guidecarnotsecond law

Two identical blocks each have heat capacity $5000$ J/K, one at $400$ K and one at $300$ K. A reversible engine runs between them until they reach a common temperature. Assume the heat capacities do not change with temperature. How much work does the engine extract in total, in J?

31.

Otto cycle measured against Carnot

Very HardWritten for this guideotto cyclecarnot

An ideal Otto cycle with compression ratio $10.0$ and $\gamma = 1.40$ has a lowest temperature of $300$ K and a highest temperature of $2000$ K. A Carnot engine between those same extreme temperatures would be the best possible. What percentage of the Carnot efficiency does the Otto cycle achieve, in percent?

Also worth doing

Problems from past exams and textbooks, cited rather than reproduced.

USAPhO 2018 Problem B1

HardPast examUSAPhO 2018cycles

Official exam text is not reproduced here. Open the USAPhO 2018 exam (PDF)

Solutions

1. Carnot ceiling Answer: 50 %

$\eta_{max} = 1 - T_C/T_H = 1 - 0.5 = 50\%$ . Any real engine between these reservoirs does worse; claiming better violates the second law.

2. Efficiency from work and heat input Answer: 30 %

Thermal efficiency is $e = W/Q_H$ . Here $e = 300/1000 = 0.300$ , or $30\%$ .

3. Carnot efficiency between two reservoirs Answer: 40 %

For a Carnot engine $e = 1 - T_C/T_H$ with absolute temperatures. So $e = 1 - 300/500 = 0.400$ , or $40\%$ .

4. Heat dumped to the cold reservoir Answer: 550 J

Over a full cycle $\Delta U = 0$ , so energy conservation gives $Q_C = Q_H - W$ . Thus $Q_C = 800 - 250 = 550$ J.

5. Coefficient of performance of a fridge Answer: 4

For a refrigerator the useful output is the heat pulled from the cold side: $\mathrm{COP} = Q_C/W$ . So $\mathrm{COP} = 400/100 = 4$ .

6. Coefficient of performance of a heat pump Answer: 4

For a heat pump the useful output is the heat delivered indoors: $\mathrm{COP} = Q_H/W = 600/150 = 4$ .

7. Best possible fridge COP Answer: 9

The best case is a reversed Carnot cycle, with $\mathrm{COP}_{max} = T_C/(T_H - T_C)$ . So $\mathrm{COP}_{max} = 270/30 = 9$ .

8. Power output from cycles per second Answer: 10000 W

Power is work per cycle times cycle rate: $P = Wf = (500)(20.0) = 10000$ W.

9. Net work from heat in and out Answer: 500 J

Since the engine returns to its initial state, $W = Q_H - Q_C = 2000 - 1500 = 500$ J.

10. Required hot reservoir temperature Answer: 400 K

From $e = 1 - T_C/T_H$ we get $T_H = T_C/(1-e)$ . So $T_H = 300/0.750 = 400$ K.

11. Otto cycle efficiency Answer: 56.47 %

The Otto cycle efficiency is $e = 1 - r^{1-\gamma}$ . With $r = 8.00$ and $\gamma = 1.40$ , $e = 1 - 8.00^{-0.400} = 56.47\%$ once converted.

12. Diesel cycle efficiency Answer: 63.16 %

Substitute directly: $r_c^{1.40} = 2.639$ , $r^{0.400} = 3.178$ , so the correction term is $(1/1.40)(1.639)/(3.178\times1.00) = 0.3683$ . Then $e = 1 - 0.3683 = 63.16\%$ .

13. Brayton cycle for a gas turbine Answer: 48.21 %

The exponent is $(1-1.40)/1.40 = -0.2857$ , so $10.0^{-0.2857} = 0.5179$ . Therefore $e = 1 - 0.5179 = 48.21\%$ .

14. Testing an inventor's claim Answer: 500 J

No engine beats Carnot: $e_{max} = 1 - 300/600 = 0.500$ . So $W_{max} = e_{max}Q_H = 500$ J, and the claimed $600$ J is impossible.

15. Second-law efficiency of a real engine Answer: 60 %

The second-law (relative) efficiency is $e/e_{Carnot}$ . Here $0.300/0.500 = 0.600$ , i.e. $60\%$ of the theoretical maximum.

16. Waste heat from a power plant Answer: 1857 MW

The fuel supplies $Q_H = W/e = 1000/0.350 = 2857$ MW of heat, and the rest is rejected: $Q_C = Q_H - W$ . So $Q_C = 2857 - 1000 = 1857$ MW.

17. Ideal Stirling engine efficiency Answer: 66.67 %

The regenerator returns the heat stored during the constant-volume cooling stage, so only the isothermal heat exchanges count and $e = 1 - T_C/T_H$ . Here $e = 1 - 300/900 = 66.67\%$ .

18. Power from fuel burn rate Answer: 20000 W

Heat input rate is $(4.00\times10^7)(2.00\times10^{-3}) = 8.00\times10^4$ W, and $P = e\dot{Q}_H$ . So $P = 0.250(8.00\times10^4) = 20000$ W.

19. Work needed by a refrigerator Answer: 166.7 J

From $\mathrm{COP} = Q_C/W$ , $W = Q_C/\mathrm{COP} = 500/3.00 = 166.7$ J.

20. Relating the two COPs Answer: 5.5

Energy conservation gives $Q_H = Q_C + W$ , so $\mathrm{COP}_{hp} = Q_H/W = Q_C/W + 1 = \mathrm{COP}_{ref} + 1$ . Thus $\mathrm{COP}_{hp} = 5.5$ .

21. Electrical power drawn by a heat pump Answer: 3429 W

Since $\mathrm{COP} = \dot{Q}_H/P$ , the input power is $P = \dot{Q}_H/\mathrm{COP} = 12000/3.50 = 3429$ W.

22. Work per cycle from efficiency Answer: 440 J

By definition $W = eQ_H$ . So $W = 0.550(800) = 440$ J.

23. Work as the area of a PV rectangle Answer: 600 J

Net work per cycle equals the enclosed area on the $PV$ diagram, positive for a clockwise loop. The area is $\Delta P\,\Delta V = (2.00\times10^5)(3.00\times10^{-3}) = 600$ J.

24. Carnot engine driving a Carnot fridge Answer: 3125 J

The engine's work is $W = (1 - 300/800)(1000) = 625$ J. The refrigerator has $\mathrm{COP} = T_C/(T_H-T_C) = 250/50 = 5.00$ , so $Q_C = 5.00W = 3125$ J.

25. Two Carnot engines in series Answer: 66.67 %

The first engine has $e_1 = 1/3$ and passes $Q_H(2/3)$ to the second, which converts half of that. Total work is $Q_H(1/3 + 1/3) = (2/3)Q_H$ , which equals $1 - 300/900 = 66.67\%$ , exactly the Carnot value for the outer pair.

26. Otto cycle with a monatomic working gas Answer: 76.89 %

The Otto result $e = 1 - r^{1-\gamma}$ holds for any ideal gas. With $\gamma - 1 = 2/3$ , $9.00^{2/3} = 4.327$ , so $e = 1 - 1/4.327 = 76.89\%$ . A larger $\gamma$ makes the adiabats steeper and the cycle more efficient.

27. Ocean thermal energy conversion Answer: 12.04 MW

The Carnot efficiency is $e = 1 - 277/300 = 0.0767$ , so the required heat input is $\dot{Q}_H = 1.00/e = 13.04$ MW. The rejected rate is $\dot{Q}_H - 1.00 = 12.04$ MW: tiny temperature differences demand enormous water flow.

28. By how much does the claim break the second law? Answer: 7.5 percentage points

The Carnot limit is $e_{max} = 1 - 250/400 = 37.5\%$ . Coupling the claimed engine to a reversed Carnot engine would move heat from cold to hot with no other effect, which the Clausius statement forbids. The excess is $45.0 - 37.5 = 7.5$ percentage points.

29. Net work of a Stirling cycle Answer: 1728 J

The constant-volume legs do no work, so $W = nRT_H\ln 2 - nRT_C\ln 2 = nR(T_H-T_C)\ln 2$ . That is $(1.00)(8.31)(300)(0.6931) = 1728$ J.

30. Work from two finite bodies Answer: 35900 J

Reversibility requires the total entropy change $C\ln(T_f/400) + C\ln(T_f/300)$ to vanish, giving $T_f = \sqrt{(400)(300)} = 346.4$ K. Energy conservation then gives $W = C(400 + 300 - 2T_f) = 5000(700 - 692.8) = 3.59e+04$ J.

31. Otto cycle measured against Carnot Answer: 70.81 %

The Otto efficiency is $1 - 10.0^{-0.400} = 0.6019$ , while the Carnot value for the same temperature extremes is $1 - 300/2000 = 0.8500$ . The ratio is $70.81\%$ : the Otto cycle loses ground because it absorbs and rejects heat over ranges of temperature, not at the two extremes only.