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Grade 10-11 (age 15-17)

Heat Engines & Efficiency

Heat Engines & Efficiency

Only Part of the Heat Becomes Work
300 J
👀 See It
①A heat engine absorbs heat Q₁ from the hot reservoir
②Only part becomes work W; the rest Q₂ is dumped to the cold reservoir
③Less dumped heat means higher efficiency — but it can never reach zero
Work Done and Efficiency
Work done by the engine
W = Q1 − Q2
The work output equals heat absorbed minus heat dumped
Thermal efficiency
e = WQ1 = 1 − Q2Q1
The fraction of absorbed heat turned into work — a value between 0 and 1
The Second Law of Thermodynamics
🚫 Efficiency Can Never Be 1
①The dumped heat Q₂ cannot be made 0 (second law)
②Hence the efficiency e is always less than 1
③No engine turns 100% of absorbed heat into work
Compute It Directly
Example 1
A heat engine absorbs 400 J and dumps 300 J. Find the work done and the efficiency.
1
The work 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 absorbed heat Q₁ — not by the dumped heat Q₂.
Example 2
An engine with efficiency 0.3 absorbs 600 J. Find the work done and the heat dumped.
1
Find the work from W = e·Q₁.
W = 0.3 × 600 = 180 J
2
The dumped heat is Q₂ = Q₁ − W.
Q2 = 600 − 180 = 420 J
▸ W = 180 J, Q₂ = 420 J
Knowing efficiency and absorbed heat gives the work and dumped heat at once.
Wrap-up
Key result
W = Q1 − Q2, e = WQ1 = 1 − Q2Q1 < 1
Work is absorbed minus dumped; efficiency is work over absorbed — always below 1
2020 CSAT Science (Physics Ⅰ) type, adapted
An engine with 40% efficiency does 200 J of work per cycle. How much heat did it absorb from the hot reservoir?
①300 J
②400 J
③500 J
④600 J
⑤800 J
▸ ③ 500 J
1
From e = W/Q₁, we get Q₁ = W/e.
Q1 = We = 2000.4
2
Compute.
Q1 = 500 J
🎯 Exam Points
①W = Q₁ − Q₂ (absorbed − dumped)
②e = W/Q₁ = 1 − Q₂/Q₁
③divide by absorbed heat (not dumped)
④second law: e < 1 (100% impossible)
⑤knowing two of W, Q₁, Q₂, e fixes the rest
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