A heat engine takes heat Q₁ from a hot reservoir, turns only part of it into work W, and releases the rest, Q₂, to a cold reservoir.
So the work done is W = Q₁ − Q₂, and the efficiency e = W/Q₁, the share of the heat taken in that becomes work, is always less than 1.
This is because, by the second law of thermodynamics, the heat released can never be made 0.
Move the slider for the released heat Q₂ and the three arrows, whose widths are proportional to energy, change along with the efficiency.
Only Part of the Heat Becomes Work
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A heat engine taking Q₁ from the hot reservoir, doing work W and releasing Q₂ to the cold reservoir; arrow widths are proportional to energy
300 J
👀 Reading the figure
①The heat engine takes in heat Q₁ from the hot reservoir
②It turns only part of it into work W and releases the rest, Q₂, to the cold reservoir
③The widths of the two outgoing arrows add up to the width of the Q₁ arrow
④The less heat released, the higher the efficiency; even at the slider's smallest value, 40 J, the efficiency does not reach 1
Work and Efficiency
Work done by the engine
W = Q1 − Q2
The engine does work equal to the heat taken in minus the heat released
Thermal efficiency
e = WQ1 = 1 − Q2Q1
The share of the heat taken in that becomes work, a value between 0 and 1
The Second Law of Thermodynamics
🚫 The efficiency cannot be 1
①The heat released Q₂ cannot be made 0 (second law of thermodynamics)
②So the efficiency e is always less than 1
③No engine turns 100% of the heat it takes in into work
④The temperatures of the hot and cold reservoirs set an upper limit on the efficiency
Worked Examples
Example 1
A heat engine takes in 400 J from the hot reservoir and releases 300 J to the cold reservoir. Find the work done and the efficiency.
1
The work done is W = Q₁ − Q₂.
W = 400 − 300 = 100 J
2
The efficiency is e = W/Q₁.
e = 100400 = 0.25
▸ W = 100 J, e = 0.25 (25%)
Divide by the heat taken in, Q₁; dividing by the heat released, Q₂, gives the wrong answer.
Example 2
An engine with efficiency 0.3 takes in 600 J. Find the work done and the heat released.
1
Find the work as W = e × Q₁.
W = 0.3 × 600 = 180 J
2
The heat released is Q₂ = Q₁ − W.
Q2 = 600 − 180 = 420 J
▸ W = 180 J, Q₂ = 420 J
Knowing the efficiency and the heat taken in gives the work and the released heat at once.
Summary
Key results
W = Q1 − Q2, e = WQ1 = 1 − Q2Q1 < 1
The work is the heat taken in minus the heat released, and the efficiency is that work divided by the heat taken in, always below 1
2020 CSAT Science (Physics I) type, adapted
An engine with 40% efficiency does 200 J of work in one cycle. How much heat did it take in from the hot reservoir?
①300 J
②400 J
③500 J
④600 J
⑤800 J
▸ ③ 500 J
1
From e = W/Q₁, Q₁ = W/e.
Q1 = We = 2000.4
2
Substitute the values and compute.
Q1 = 500 J
🎯 Exam Points
①W = Q₁ − Q₂: heat taken in minus heat released
②e = W/Q₁ = 1 − Q₂/Q₁
③Divide by the heat taken in, not the heat released
④By the second law, e < 1
⑤Knowing any two of W, Q₁, Q₂ and e fixes the rest