Question

For a heat engine performing between the temperature 273 K and 409.5 K, which among the following cannot be the efficiency?​

Answer 1

Answer:

For a heat engine performing between the temperature 273 K and 409.5 K, which among the following cannot be the efficiency?

Answer 2

Concept:

• Heat engines
• Carnot cycle
• Thermodynamics

Given:

• Higher temperature T₁ = 409.5K
• Lower temperature T₂ = 273K

Find:

• The efficiency
• The value that cannot be efficiency

Solution:

The efficiency of the engine = 1-T₂/T₁

T₂/T₁ = 273/409.5 = 2/3

The efficiency of the engine = 1-2/3

The efficiency of the engine = 1/3

This value is the maximum value of the efficiency.

1/2 is greater than 1/3

1/2 cannot be the efficiency of the heat engine.

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