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 :

Carnot engine (application of the second law of thermodynamics) - problems and solutions 1

Read :  Equilibrium of the bodies on a inclined plane – application of the Newton's first law problems and solutions

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 :

Carnot engine (application of the second law of thermodynamics) - problems and solutions 2

Work is done by the Carnot engine :

W = e QH

W = (0.6)(6000)

W = 3600 Joule

Read :  Collisions – problems and solutions

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 :

Carnot engine (application of the second law of thermodynamics) - problems and solutions 3

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

Read :  Applications of Bernoulli's principle

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

Carnot engine (application of the second law of thermodynamics) - problems and solutions 4Known :

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 :

Carnot engine (application of the second law of thermodynamics) - problems and solutions 5

Work is done by the Carnot engine :

W = e QH

W = (0.625)(1000)

W = 625 Joule