# Carnot engine (application of the second law of thermodynamics) – problems and solutions

1. An engine operates between 1200 Kelvin and 300 Kelvin. What is the efficiency of this engine?

Known :

High temperature (TH) = 1200 K

Low temperature (TL) = 300 K

Wanted: Efficiency (e)

Solution :

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2. An engine operates between 727oC and 127oC. The engine’s heat input is 6000 Joule. What is the efficiency of the engine and work done by the engine each cycle?

Known :

High temperature (TH) = 727oC + 273 = 1000 K

Low temperature (TL) = 127oC + 273 = 400 K

Heat input (QH) = 6000 Joule

Wanted: Efficiency (e) and work done by the engine (W)

Solution :

The efficiency of the Carnot engine :

Work is done by the Carnot engine :

W = e QH

W = (0.6)(6000)

W = 3600 Joule

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3. An engine operates between 527oC and 127oC. The engine’s heat input is 10,000 Joule. How much heat is discharged as waste heat from this engine?

Known :

High temperature (TH) = 527oC + 273 = 800 K

Low temperature (TL) = 127oC + 273 = 400 K

Heat input (QH) = 10,000 Joule

Wanted : Heat output (QL)

Solution :

Efficiency of the Carnot engine :

Work is done by Carnot engine :

W = e Q1

W = (0.5)(10.000)

W = 5000 Joule

Heat output (QL) = Heat input (QH) – Work (W)

QL = QH – W

QL = 10,000 – 5,000

QL = 5000 Joule

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4. What is the efficiency of the engine and work done by the engine according to diagram below.

Known :

High temperature (TH) = 800 K

Low temperature (TL) = 300 K

Heat input (QH) = 1000 Joule

Wanted: Efficiency (e) and work (W) done by the engine

Solution :

The efficiency of the Carnot engine :

Work is done by the Carnot engine :

W = e QH

W = (0.625)(1000)

W = 625 Joule