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The second law, entropy $\Delta S = Q/T$, heat engine efficiency, the Carnot limit, and refrigerators and heat pumps.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to find engine efficiencies and their limits and use entropy to tell which processes can happen.
From the last two lessons you know the first law, $\Delta U = Q + W$, and that a gas taken around a cycle does net work equal to the enclosed area while its internal energy returns to its starting value. The first law says energy is conserved; it does not say which way things happen. This lesson adds the law that does.
| Term | What it means |
|---|---|
| Second law | Heat flows spontaneously only from hot to cold; total entropy never decreases. |
| Entropy | A measure of how spread out energy is; $\Delta S = Q/T$ for heat at absolute temperature $T$. |
| Heat engine | A device that turns part of the heat it absorbs into work, rejecting the rest. |
| Efficiency | $e = W/Q_H$, the fraction of heat input turned into work. |
| Carnot efficiency | $1 - T_C/T_H$, the most any engine between two temperatures can achieve. |
| Coefficient of performance | For a refrigerator, heat removed per joule of work, $Q_C/W$. |
A cup of hot coffee cools; the room never spontaneously heats the coffee. Energy would be conserved either way, so the first law cannot explain the direction. The second law does: heat flows spontaneously only from hot to cold.
Entropy makes this precise. When heat $Q$ enters a body at absolute temperature $T$, its entropy rises by
$$\Delta S = \frac{Q}{T}.$$
In any spontaneous process the total entropy of an isolated system increases. Heat leaving a hot body lowers its entropy by $Q/T_H$; the same heat entering a cold body raises its entropy by the larger amount $Q/T_C$. For engines this sets a hard limit: no engine between temperatures $T_H$ and $T_C$ can exceed the Carnot efficiency $1 - T_C/T_H$.
Another way: picture
Picture a new deck of cards in order. Shuffle it and it becomes disordered; shuffle again and it stays disordered. It could in principle return to order, but there are vastly more disordered arrangements than ordered ones. Energy spreading from hot to cold is like that shuffle: far more ways for energy to be spread out than concentrated.
Another way: steps
A heat engine takes heat $Q_H$ from a hot source, such as burning fuel or steam, turns part of it into work $W$, and rejects the rest $Q_C$ to a cold sink, such as the air or a river. Over each cycle the engine returns to its starting state, so by the first law $Q_H = W + Q_C$.
Its efficiency is the useful fraction, $e = W/Q_H$. A car engine turns about a quarter to a third of its fuel's energy into work; the rest leaves through the radiator and exhaust. A modern combined-cycle natural gas plant reaches about sixty percent, among the best of any heat engine.
In 1824 the French engineer Sadi Carnot asked how efficient an engine could possibly be. His answer, sharpened later with entropy, is that the best engine between temperatures $T_H$ and $T_C$ has efficiency $1 - T_C/T_H$, whatever its design or working substance.
The limit comes from entropy. The heat $Q_H$ taken in lowers the source's entropy by $Q_H/T_H$. The engine returns to its starting state, so for total entropy not to fall, the cold sink must gain at least $Q_H/T_H$, which means receiving at least $Q_C = Q_H T_C/T_H$. Only the rest can be work.
The Carnot formula shows the way to better engines: make $T_H$ as high as possible or $T_C$ as low as possible. The cold side is usually stuck near the temperature of the air or a river, about $300$ K. The hot side is limited by materials that can survive it.
That is why jet engine turbine blades are made of nickel superalloys with internal cooling channels and ceramic coatings, running in gas hotter than their own melting point, and why power plants seek ever-higher steam temperatures. Each hundred degrees gained on the hot side raises the efficiency limit by several percent.
A refrigerator runs an engine backward: it uses work $W$ to move heat $Q_C$ out of a cold space and releases $Q_H = Q_C + W$ into a warm one. Heat does flow from cold to hot, but not by itself; the work pays for it, and total entropy still rises.
Its coefficient of performance, $\text{COP} = Q_C/W$, can be greater than one: a typical kitchen refrigerator moves three joules of heat for each joule of electricity. The ideal limit is $T_C/(T_H - T_C)$, larger when the two temperatures are close, which is why heat pumps work best in mild winters.
Checking an answer. No efficiency can exceed the Carnot value, and none can reach $100$ percent. Total entropy must rise in a real process and stay the same only in an ideal reversible one. Heat flowing from hot to cold must give a positive total.
The formula $\Delta S = Q/T$ applies when heat enters a body at a steady temperature, as for a large reservoir or a substance changing phase. When temperature changes during heating, the entropy change is the sum of small $Q/T$ contributions, which requires calculus.
The second law itself is a statement of probability. A gas has vastly more arrangements with energy spread out than concentrated, so random molecular motion carries it toward spread-out states. For the enormous numbers of molecules in everyday objects, the odds against a decrease are so great that it never happens.
Ludwig Boltzmann showed in the 1870s that entropy counts arrangements: $S = k_B \ln \Omega$, where $\Omega$ is the number of microscopic arrangements consistent with what we observe. His equation is carved on his tombstone in Vienna.
Melting ice illustrates it. In a crystal, water molecules are locked in place; in the liquid they tumble freely, with far more possible arrangements. Melting absorbs heat at $273$ K and raises entropy by $Q/T$. A kilogram of ice gains about $1220$ J/K, a precise measure of how much freer its molecules have become.
The laws of mechanics work equally well forward and backward in time: a film of two billiard balls colliding looks fine run in reverse. Yet a film of an egg breaking run backward is obviously wrong. The difference is entropy.
Every irreversible process, from friction to mixing to heat flow, raises total entropy, and that increase defines the direction of time. The universe began in a state of remarkably low entropy, and it has been increasing ever since. Physicists still debate why the beginning was so ordered, and the question sits at the meeting point of thermodynamics and cosmology.
Because no engine can convert all its heat to work, every thermal power plant rejects more heat than electricity it makes, often nearly twice as much. That heat must go somewhere: into a river, a lake, the sea or the air through cooling towers.
The giant hourglass-shaped towers seen at many American plants evaporate water to carry heat away. Plants on rivers such as the Ohio, Mississippi and Tennessee draw billions of gallons a day for cooling, and in hot summers, when river water is already warm, some must cut output to protect fish. The second law shapes where power plants are built.
Not every way of making work is limited by Carnot. Hydroelectric turbines, wind turbines, electric motors and fuel cells convert energy without first turning it into heat, so the Carnot limit does not apply to them. A large hydroelectric turbine at Grand Coulee Dam converts over ninety percent of the water's energy.
Batteries and electric motors are why electric cars use energy so efficiently: about ninety percent of the battery's energy reaches the wheels, against thirty percent of gasoline's energy in a conventional car. The second law still applies, but these devices avoid the costly step of heat. Even so, the power plants that charge the batteries are often heat engines themselves, so the whole chain's efficiency depends on how the electricity was made.
A large American coal or nuclear plant delivers about $1000$ MW of electricity with an efficiency near $33$ to $37$ percent. The second law forces it to reject the rest: roughly $1800$ MW of heat, around the clock. A plant on a river may pump a million gallons of water a minute through its condensers, warming it by several degrees.
Where rivers are small, plants use cooling towers, which evaporate part of the water to carry heat into the air. During heat waves, when river water is already warm, the cold reservoir is warmer, the Carnot limit falls, and environmental rules on discharge temperature can force plants such as those on the Tennessee River to reduce output just when demand for air conditioning peaks.
Heat pumps are replacing furnaces across the United States, encouraged by federal tax credits. Instead of burning fuel, they use electricity to move heat from outdoor air into the house. A good unit delivers three to four joules of heat for each joule of electricity at $0$ °C.
The second law explains their weakness: as outdoor air gets colder, the gap between inside and outside grows, and the ideal coefficient of performance $T_H/(T_H - T_C)$ falls. At $-20$ °C a basic heat pump might manage only $1.5$. Cold-climate models tested in Maine and Minnesota use variable-speed compressors and refrigerants chosen to keep working well below zero, and they are now common even in northern New England.
It is natural to think that with enough engineering an engine could turn all its heat into work. The second law forbids it: every heat engine must reject some heat to a colder reservoir, and even an ideal one is limited to $1 - T_C/T_H$. This is not a matter of friction or leaks, which only make things worse.
A related error is to think refrigerators destroy heat or that a refrigerator with its door open cools a room. It moves heat from inside to the coils behind it, and since it adds its own work, an open refrigerator actually warms the kitchen.
A turbine uses steam at $550$ °C and condenses it at $40$ °C. Convert to kelvins.
$T_H = 823\ \text{K}, \quad T_C = 313\ \text{K}$
Absolute temperatures.
Find the Carnot limit.
$e_{\max} = 1 - \dfrac{313}{823} = 0.620$
At best $62$ percent.
The real turbine achieves $42$ percent. For $1000$ MW of heat in, find the work.
$W = 0.42 \times 1000 = 420\ \text{MW}$
Mechanical power out.
Find the heat rejected.
$Q_C = 1000 - 420 = 580\ \text{MW}$
To the condenser.
Find the entropy made per second.
$\dfrac{580}{313} - \dfrac{1000}{823} = 1.853 - 1.215 = 0.638\ \text{MW/K}$
Positive, as the second law requires.
$500$ J of heat leaks from a room at $300$ K to a freezer interior at $250$ K. Find the room's entropy change.
$\Delta S_{\text{room}} = -\dfrac{500}{300} = -1.667\ \text{J/K}$
It loses heat.
Find the freezer's entropy change.
$\Delta S_{\text{freezer}} = +\dfrac{500}{250} = +2.000\ \text{J/K}$
It gains the heat.
Add the two changes.
$\Delta S = 2.000 - 1.667 = 0.333\ \text{J/K}$
A net increase.
Consider the reverse flow.
$\Delta S = -0.333\ \text{J/K}$
Forbidden without work.
Find the least work to pump it back.
$W = \dfrac{500}{\text{COP}_{\max}} = \dfrac{500}{5} = 100\ \text{J}$
The ideal value is five here.
Interpret the result.
$\text{leaks cost the freezer electricity}$
Good seals save energy.
A heat pump warms a house at $293$ K from outdoor air at $268$ K. Find the ideal heating performance.
$\dfrac{Q_H}{W} = \dfrac{T_H}{T_H - T_C} = \dfrac{293}{25} = 11.7$
Heat delivered per joule of work.
A real unit achieves a third of that. Find its performance.
$\dfrac{11.7}{3} = 3.9$
Still well above one.
The house needs $10$ kW of heat. Find the electric power.
$W = \dfrac{10}{3.9} = 2.56\ \text{kW}$
Compared with $10$ kW for electric heaters.
Find the heat drawn from outdoors.
$Q_C = 10 - 2.56 = 7.44\ \text{kW}$
Free energy from cold air.
Repeat the ideal value for $248$ K outdoors.
$\dfrac{293}{45} = 6.5$
Colder air, lower performance.
Explain why heat pumps struggle in severe cold.
$\text{a larger } T_H - T_C \text{ costs more work}$
Cold-climate models use better compressors.
Write the formula.
$e_{\max} = 1 - \dfrac{T_C}{T_H}$
Both in kelvins.
Substitute the temperatures.
$1 - \dfrac{400}{600}$
The fraction rejected.
Evaluate the limit.
An engine takes in heat from steam at $400$ °C and rejects heat to cooling water at $40$ °C. What is the greatest efficiency it could possibly have?
Complete the worked solution: a refrigerator removes $1200$ J from its interior at $270$ K using $400$ J of electrical work, and releases heat into a kitchen at $310$ K. Find the heat released in J, the coefficient of performance, and the coefficient of performance of an ideal refrigerator between the same temperatures.
Balance the energy.
$Q_H = Q_C + W =$ r
All the energy ends up in the kitchen.
Find the coefficient of performance.
$\text{COP} = \dfrac{Q_C}{W} =$ p
Heat removed per joule of work.
Find the ideal limit.
$\text{COP}_{\max} = \dfrac{T_C}{T_H - T_C} =$ i
A Carnot refrigerator.
Compare the two.
$\text{COP} < \text{COP}_{\max}$
Real machines fall short.
Match each idea to its statement.
| heat flows spontaneously only from hot to cold | no cyclic engine turns heat entirely into work | heat added divided by absolute temperature | the highest efficiency between two temperatures | |
|---|---|---|---|---|
| Clausius statement | ||||
| Kelvin–Planck statement | ||||
| entropy change | ||||
| Carnot efficiency |
Each cycle, an engine absorbs $1200$ J from a reservoir at $700$ K and does $420$ J of work, rejecting the rest to a reservoir at $350$ K. Fill in the heat rejected in J, the engine's efficiency in percent, and the Carnot efficiency in percent.
| value | |
|---|---|
| heat rejected (J) | |
| efficiency (%) | |
| Carnot efficiency (%) |
Heat $Q$, in joules, leaks from a large reservoir at $500$ K to another at $400$ K, so small that neither temperature changes. Write the total entropy change of the two reservoirs, in J/K, as a function of $Q$.
Answer:
$0.50$ kg of ice at $0$ °C melts into water at $0$ °C. The latent heat of fusion is $334$ kJ/kg. By how much does its entropy increase, in J/K?
Answer: J/K
A power plant on the Ohio River delivers $1200$ MW of electricity with an overall efficiency of $38$ percent. At what rate, in MW, does it reject heat to the river and the air?
Answer: MW
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Heat $Q$, in joules, leaks from a large reservoir at $800$ K to another at $400$ K, so small that neither temperature changes. Write the total entropy change of the two reservoirs, in J/K, as a function of $Q$.
Answer:
You can apply the second law. Explain to someone why a power plant must reject heat even if it were built perfectly.
21. Your turn: an engine runs between $600$ K and $400$ K. What is its Carnot efficiency?, step 3
$e_{\max} = 0.333$
At most a third.