SSC JE Mechanical Engineering 27 Oct 2020 Shift 1 Papers

Question : 143 of 200

Marks: +1, -0

If the temperature of the heat source is 1990 K and the sink is 850 K, then what is the Carnot efficiency

Solution:

Given:
The temperature of heat source,

T_H

= 1990 K, The temperature of the heat sink,

T_L

= 850 K
Carnot efficiency

⇒ \η = 1 - {{{T_L}}}/{{{T_H}}} = 1 - {{850}}/{{1990}} = 0.573

η = 0.573 or 57.3 %

Go to Question:

Prev Question

Next Question