high school physics Heat Engines & Efficiency

Heat Engines & Efficiency

A heat engine absorbs heat Q₁ from a hot reservoir, turns only part into work W, and dumps the rest Q₂ to a cold reservoir. Work is W = Q₁ − Q₂ and efficiency e = W/Q₁ is always less than 1. The second law of thermodynamics forbids dumping heat of zero. Change the dumped heat Q₂ and work and efficiency change.

Only Part of the Heat Becomes Work

Heat dumped Q₂ (Q₁=400 fixed, 40~400)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. The slider is 40~400 so e does not hit 1

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

④Hot and cold temperatures cap the efficiency. Slider floor is 40

Compute It Directly

Example 1

A heat engine absorbs 400 J and dumps 300 J. Find the work done and the efficiency.

The work is W = Q₁ − Q₂.

W = 400 − 300 = 100 J

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.

Find the work from W = e·Q₁.

W = 0.3 × 600 = 180 J

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

exam-style

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

From e = W/Q₁, we get Q₁ = W/e.

Q₁ = We = 2000.4

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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Heat and Laws of Thermodynamics

Properties of Waves

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