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 (T_H) = 1200 K

Low temperature (T_L) = 300 K

Wanted: Efficiency (e)

Solution :

[irp]

2. An engine operates between 727^oC and 127^oC. 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 (T_H) = 727^oC + 273 = 1000 K

Low temperature (T_L) = 127^oC + 273 = 400 K

Heat input (Q_H) = 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 Q_H

W = (0.6)(6000)

W = 3600 Joule

[irp]

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

Known :

High temperature (T_H) = 527^oC + 273 = 800 K

Low temperature (T_L) = 127^oC + 273 = 400 K

Heat input (Q_H) = 10,000 Joule

Wanted : Heat output (Q_L)

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 (Q_L) = Heat input (Q_H) – Work (W)

Q_L = Q_H – W

Q_L = 10,000 – 5,000

Q_L = 5000 Joule

[irp]

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

Known :

High temperature (T_H) = 800 K

Low temperature (T_L) = 300 K

Heat input (Q_H) = 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 Q_H

W = (0.625)(1000)

W = 625 Joule