Physics 2nd Paper · Classes 11-12 · Chapter 1 · 1.10-1.11

Heat Engine and Refrigerator.

An engine takes heat from something hot and gives you work. A fridge spends work to pull heat out of something cold. It is the same picture with the arrows turned round. The two formulas here turn up in almost every board paper, and both hide the same trap: the temperatures must be in kelvin.

1. Work out of heat

To get work out of heat you need a mechanical arrangement. It has a heat source and a heat sink, and the source has to be hotter than the surroundings.

A device that turns heat energy into mechanical energy is a heat engine. Steam engines, petrol engines and diesel engines are all heat engines.

The principle in one line: the working substance takes heat Q₁ from a hot source, turns part of it into work and throws the rest, Q₂, into a cold sink. To keep the work coming it runs in a cycle. The work is W = Q₁ − Q₂.

The bigger the share of its heat an engine turns into work, the more efficient it is. By that measure a petrol engine beats a steam engine.

2. What efficiency means

The efficiency of an engine is the heat turned into work divided by the heat taken in: η = (heat turned into work)/(heat absorbed) = (Q₁ − Q₂)/Q₁.

Split the fraction and the thermal efficiency reads η = (Q₁ − Q₂)/Q₁ = 1 − Q₂/Q₁ (1.24). As a percentage, η = (1 − Q₂/Q₁) × 100%.

The less heat thrown away as Q₂, the higher η. But no heat engine turns all its heat into work; an ordinary one manages about 30%.

Suppose no heat at all were thrown away, Q₂ = 0. Then W = Q₁ and η = 1, which is 100%. The sum works on paper; no real engine can do it.

3. Efficiency from temperatures

In an ideal engine the heat taken in and thrown out is proportional to the temperature of the source and of the sink: Q₁/T₁ = Q₂/T₂, so Q₂/Q₁ = T₂/T₁.

Put that into the earlier formula and η = (T₁ − T₂)/T₁ = 1 − T₂/T₁, or as a percentage η = (T₁ − T₂)/T₁ × 100% (1.25). Here T₁ and T₂ are in kelvin.

That word “kelvin” costs more marks than anything else here. Convert first, T = t + 273: 127°C and 27°C give T₁ = 400 K and T₂ = 300 K, so η = 1 − 300/400 = 25%.

One engine, worked two ways: once with Celsius put straight in, once in kelvin. Change the temperatures and see how big the mistake is.

Now you can see why efficiency never reaches 100%. η = 100% needs T₂ = 0, and no body can be cooled to 0 K. So even the ideal Carnot engine is not 100% efficient.

Turn it round: cool the sink and a Carnot engine gets more efficient, because as T₂ falls, (T₁ − T₂) grows. Keep the source at 600 K and bring T₂ down.

The Carnot engine is the ideal engine. Its efficiency is the heat turned into work in one cycle over the heat taken from the source, and it is higher than that of any practical engine.

Only T₁ and T₂ appear in the formula, so the efficiency depends only on the temperatures of the source and the sink. What the working substance is makes no difference.

For the same reason, all reversible engines working between the same two temperatures have the same efficiency.

The wider the gap between source and sink, the higher the efficiency. Yet T₁ is always bigger than (T₁ − T₂), so the efficiency never gets to 100%.

4. The book's four examples

(1) An engine throws away 70% of the heat it takes in. Then Q₂ = 0.7 Q₁ and η = (Q₁ − 0.7Q₁)/Q₁ × 100% = 30%.

(2) It takes 840 J from a 400 K source and rejects 630 J. η = (840 − 630)/840 × 100% = 25%. Then 0.25 = (400 − T₂)/400 gives T₂ = 300 K.

(3) The sink is at 27°C, which is 300 K, and η = 50%. 0.5 = 1 − 300/T₁ gives T₁ = 600 K. For η = 60%, T₁ = 750 K. So the source must be raised by 150 K.

(4) A Carnot engine between 27°C and 160°C: η = (433 − 300)/433 × 100% = 30.7%. Supply 8.4 × 10⁴ J and W = ηQ = 0.307 × 8.4 × 10⁴ ≈ 25788 J. Without rounding η first it comes to 25801 J.

5. The refrigerator: the machine in reverse

Sending heat from a lower temperature up to a higher one is called refrigeration.

A device that makes a temperature lower than its surroundings and holds it there is a refrigerator. It keeps fish, meat, drinking water and ice cream.

A refrigerator is a heat engine run backwards. It removes heat Q₁ from a cold body and dumps Q₂ into a hot one. That takes mechanical work W, so Q₂ = Q₁ + W.

In the fridge at home, Q₁ is the heat the evaporator coil pulls out of the inside, W is the work of the compressor motor, and Q₂ is the heat let out into the room air at the back.

The heat removed divided by the compressor's work is the coefficient of performance, K: K = Q₁/W = Q₁/(Q₂ − Q₁).

A fridge pulls 500 J of heat out of its inside. Change how much work the compressor does and see where K goes.

The larger K, the more heat is moved for less work. An ordinary refrigerator has K between 2 and 6.

In temperatures, a refrigerator's efficiency is η = Q₁/W ≤ T₁/(T₂ − T₁). Watch the roles: here T₁ is the cold side, where heat is taken from, and T₂ the hot side, where it goes.

Put the two side by side. An engine takes heat from a hot source, does work and lets the unused heat go to a cold sink. A refrigerator takes heat out of a cold body and lets it go into a hot one, and it needs energy from outside to do it. Wipe across and watch the arrows turn.

The stimulus of a CQ is usually exactly this block diagram. Now build it yourself: put each label in its place, first for the engine, then for the fridge.

6. Two refrigerator examples

(1) K = 4.6 and Q₁ = 250 J. Then W = Q₁/K = 250/4.6 ≈ 54 J, and Q₂ = W + Q₁ = 250 + 54 = 304 J.

(2) It takes 3500 J out of a body at −70°C (203 K) and rejects 5200 J. K = 3500/(5200 − 3500) = 2.06. Then from K = T/(T_hot − T), with T the cold 203 K, the hot side comes to about 301.6 K = 28.6°C. The book swaps the Q and T subscripts in this example, so go by which side is cold and which is hot, not by the labels.

(3) With K = 2.5 and Q₁ = 500 J, the work supplied is W = Q₁/K = 500/2.5 = 200 J and the heat rejected is Q₂ = 700 J.

These are the two formulas people mix up most. Touch every cell of the table.

7. Board sums

An engine takes 500 J from a source at 327°C (600 K) and rejects heat to a sink at 27°C (300 K). η = 1 − 300/600 = 50%, and W = 0.5 × 500 = 250 J.

If that engine's sink warms to 177°C (450 K), η = 1 − 450/600 = 25%. The two efficiencies are in the ratio 50 : 25 = 2 : 1.

An engine takes 1200 J from a 600 K source and rejects 300 J. From Q₁/T₁ = Q₂/T₂, T₂ = T₁ × Q₂/Q₁ = 600 × 300/1200 = 150 K, and η = 1 − 300/1200 = 75%.

An engine between 227°C and 27°C. Warm the source a little, or cool the sink a little? Try both by the same amount.

An engine runs between 27°C and 227°C: η = 1 − 300/500 = 40%. Raising the source by 20°C on its own (520 K) gives η = 1 − 300/520 = 42.3%; lowering the sink by 20°C instead (280 K) gives η = 1 − 280/500 = 44%. Both raise the efficiency.

Every sum in this part is one of the seven kinds below. Type what you know and leave the unknown box empty.

8. Check yourself

Three questions in the exam's style. Keep kelvin in mind for the first.

For the second, think about what happens if nothing is thrown away.

The last is a board sum, like the ones in section 7.

The whole note on eight cards.

Source: Physics 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Thermodynamics”, section 1.10-1.11.