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Thermodynamics · Part 9 of 10

Reversible Processes, Heat Engines and Refrigerators

A heat engine repeats a cycle that turns heat into work; a refrigerator runs one backwards. This part starts from reversibility, then uses $W = Q_{1} - Q_{2}$ , efficiency $η = W / Q_{1}$ and the refrigerator's coefficient of performance $Q_{2} / W$ , with a worked example for each.

Builds on: Part 3 · Indicator Diagrams, State Variables and Cycles, Part 7 · Free Expansion of a Gas.

Reversible Processes, Heat Engines and Refrigerators

Video coming soon

Reversibility

Ideal vs real

A reversible process can be undone with the system and surroundings restored. It must be quasi-static and free of friction. Real processes (sudden expansion, friction, heat across a finite ΔT) are irreversible.

Table comparing reversible and irreversible processes. — Reversible processes are an ideal limit.

Heat engine

$W = Q_{1} - Q_{2}$ , $η = W / Q_{1}$

Each cycle takes $Q_{1}$ from the hot reservoir, does work W and rejects $Q_{2}$ to the cold one. ΔU = 0 over a cycle, so $W = Q_{1} - Q_{2}$ and $η = 1 - Q_{2} / Q_{1} < 1$ .

Refrigerator

$COP = Q_{2} / W$

Work W done on it moves $Q_{2}$ out of the cold space and dumps $Q_{1} = Q_{2} + W$ into the room. COP = $Q_{2} / W = Q_{2} / (Q_{1} - Q_{2})$ , usually more than 1.

Energy-flow diagram of a refrigerator. — 400 J removed with 100 J of work: 500 J to the room.

Summary

Key formulas

W = Q_{1} - Q_{2}

Heat Engines

η = \frac{W}{Q _{1}} = 1 - \frac{Q _{2}}{Q _{1}}

Thermal Efficiency of a Heat Engine

COP = \frac{Q _{2}}{W} = \frac{Q _{2}}{Q _{1} - Q _{2}}

Refrigerators

Worked examples

One for every idea

Reversible Processes

1. Is the free expansion of a gas into a vacuum a reversible process? Justify your answer.

To undo it, compress the gas back: work must be done on it.
To keep T the same, that energy leaves as heat to the surroundings.
The gas is restored, but the surroundings have lost work and gained heat.

No: free expansion is irreversible.

Heat Engines

2. A heat engine absorbs 2000 J from a hot reservoir per cycle and rejects 1200 J to a cold reservoir. Find the work done per cycle.

ΔU = 0 over a cycle: $W = Q_{1} - Q_{2} = 2000 - 1200$ .

$W = 800$ J per cycle.

Thermal Efficiency of a Heat Engine

3. For the same engine ( $Q_{1}$ = 2000 J, $Q_{2}$ = 1200 J, W = 800 J), find its thermal efficiency.

$η = W / Q_{1} = 800/2000$ .
Check: $1 - Q_{2} / Q_{1} = 1 - 0.6$ .

$η = 0.4$ , i.e. 40%.

Refrigerators

4. A refrigerator removes 400 J from its cold chamber per cycle while 100 J of work is done on it. Find its COP and the heat rejected to the surroundings.

$COP = Q_{2} / W = 400/100$ .
$Q_{1} = Q_{2} + W = 400 + 100$ .

COP = 4; $Q_{1} = 500$ J.

JEE-style question

Your turn

A heat engine of efficiency 25% does 300 J of work per cycle. The heat it rejects per cycle is:

(a)900 J

(b)1200 J

(c)75 J

(d)225 J

Show the answer and the traps

Q₁ = W/η = 300/0.25 = 1200 J, so Q₂ = Q₁ − W = 900 J: option a.

1200 J is the heat absorbed, not rejected. 75 J multiplies by η instead of dividing.

And 225 J puts (1 − η) on the work instead of on the heat.

Watch out

Common mistakes

Calling COP an efficiencyCOP = heat removed per joule of work and is usually more than 1.

Thinking an engine can reject no heatSome heat must always go to the cold reservoir (Part 10: Kelvin–Planck).

Using η = Q₂/Q₁η = W/Q₁ = 1 − Q₂/Q₁.

Practice

Try these

1. An engine of efficiency 30% absorbs 1500 J per cycle. Find the work done and the heat rejected per cycle.

W = 450 J; $Q_{2}$ = 1050 J.

2. A refrigerator with COP 5 removes 600 J per cycle from the cold space. Find the work needed and the heat rejected.

W = 120 J; $Q_{1}$ = 720 J.

3. Which is approximately reversible: (a) a block sliding to rest by friction, (b) very slow isothermal compression of a gas, (c) two gases mixing, (d) heat flowing from a hot body to a cold one?

(b).