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Thermodynamics · Part 10 of 10

Second Law and the Carnot Cycle

Engines must reject heat and refrigerators must take work, and the second law says no design can avoid it. This last part states the Kelvin–Planck and Clausius forms, then builds the Carnot cycle and its efficiency $η = 1 - T_{2} / T_{1}$ , the best any engine can do between two temperatures.

Builds on: Part 6 · Adiabatic Process, Part 9 · Reversible Processes, Heat Engines and Refrigerators.

Video coming soon

Kelvin–Planck

No 100% engine

No engine working in a cycle can take heat from one reservoir and convert all of it into work with no other effect. Some heat must go to a colder body.

Engine diagram with no cold reservoir flow, crossed out. — A machine with no heat rejected is impossible.

Refrigerator diagram with no work input, crossed out. — Cold to hot needs work.

Clausius

No free refrigerator

Heat cannot flow by itself from a colder body to a hotter one. The two statements are equivalent: a perfect engine driving a refrigerator would move heat from cold to hot with no outside work.

Carnot cycle

Two isotherms, two adiabats

AB isothermal expansion at $T_{1}$ ( $Q_{1}$ in), BC adiabatic expansion to $T_{2}$ , CD isothermal compression at $T_{2}$ ( $Q_{2}$ out), DA adiabatic compression. The adiabats make $V_{B} / V_{A} = V_{C} / V_{D}$ , so $Q_{2} / Q_{1} = T_{2} / T_{1}$ .

Carnot cycle of two isotherms and two adiabats. — Carnot cycle on a P–V diagram.

Carnot efficiency

$η = 1 - T_{2} / T_{1}$

No engine between the same reservoirs can beat Carnot, and every reversible one matches it. η depends only on the temperatures (in kelvin); η = 1 would need $T_{2} = 0$ K.

Summary

Key formulas

η_{Carnot} = 1 - \frac{T _{2}}{T _{1}}

Carnot Cycle

Worked examples

One for every idea

Statement of Second Law of Thermodynamics

1. A proposed engine absorbs 500 J of heat from a reservoir and converts all 500 J into work, rejecting no heat anywhere. Is this possible? Which statement of the second law does it violate?

It works in a cycle with $Q_{2} = 0$ : η = 100%.
The first law is satisfied (500 J in, 500 J out), but the Kelvin–Planck statement forbids it.

Impossible: it violates the Kelvin–Planck statement of the second law.

Carnot Cycle

2. A Carnot engine operates between a source at 600 K and a sink at 300 K, absorbing 1000 J per cycle. Find its efficiency and the work done per cycle.

$η = 1 - T_{2} / T_{1} = 1 - 300/600 = 0.5$ .
$W = η Q_{1} = 0.5 \times 1000$ ; the other 500 J goes to the sink.
In °C, $1 - 27/327$ would wrongly give 92%.

η = 50%, W = 500 J.

JEE-style question

Your turn

A Carnot engine with its sink at 27 °C has an efficiency of 40%. To raise its efficiency to 50% with the same sink, the source temperature must be raised by:

(a)100 K

(b)500 K

(c)600 K

(d)9 K

Show the answer and the traps

Sink 300 K. At 40%: T₁ = 300/0.6 = 500 K. At 50%: T₁ = 300/0.5 = 600 K. A rise of 100 K: option a.

500 K and 600 K are the two source temperatures, not the rise.

And 9 K comes from using 27 °C as if it were kelvin.

Watch out

Common mistakes

Using Celsius in η = 1 − T₂/T₁Always use kelvin.

Expecting a real engine to reach Carnot efficiencyCarnot is the upper limit; real (irreversible) engines do worse.

Thinking a perfect engine breaks the first lawIt conserves energy; it breaks the second law (Kelvin–Planck).

Practice

Try these

1. A Carnot engine works between 127 °C and 27 °C. Find its efficiency.

$1 - 300/400 = 25%$ .

2. A Carnot engine of efficiency 40% rejects 600 J per cycle. Find the heat absorbed and the work done.

$Q_{2} / Q_{1} = 0.6$ : $Q_{1} = 1000$ J, W = 400 J.