Question

A carnot engine operates between temperature levels of 600k and 300k, It drives a carnot refrigerator which provides cooling at 250k and discards heat at 300k. Determine a numerical value for the ratio of the heat extracted by the refrigerator (cooling load) to the heat delivered to the engine (heating load).

Answer 1

Answer:

the ratio of the heat extracted by the refrigerator (cooling load) to the heat delivered to the engine (heating load) is 2.5

Explanation:

Efficiency is defined as the ratio of desired output / input

Efficiency of carnot engine =

$n_C = \frac{W}{Q_H} \\= 1 - \frac{300}{600} \\= 0.5$

so,

$W = 0.5Q_H\\= \frac{250}{300 - 250} \\= 5$

in this case

$W = 0.2Q_L$

work is the same because we use work from carnot engine to power the refrigerator

Hence,

$0.2Q_L = 0.5Q_H\\\frac{Q_L}{Q_H} = \frac{0.5}{0.2} \\= 2.5$

the ratio of the heat extracted by the refrigerator (cooling load) to the heat delivered to the engine (heating load) is 2.5