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 =
[tex]n_C = \frac{W}{Q_H} \\= 1 - \frac{300}{600} \\= 0.5[/tex]
so,
[tex]W = 0.5Q_H\\= \frac{250}{300 - 250} \\= 5[/tex]
in this case
[tex]W = 0.2Q_L[/tex]
work is the same because we use work from carnot engine to power the refrigerator
Hence,
[tex]0.2Q_L = 0.5Q_H\\\frac{Q_L}{Q_H} = \frac{0.5}{0.2} \\= 2.5[/tex]
the ratio of the heat extracted by the refrigerator (cooling load) to the heat delivered to the engine (heating load) is 2.5
