Question

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

Answer 1

Step-by-step explanation:

Formula for maximum efficiency of a Carnot refrigerator is as follows.

`\frac{W}{Q_{H_(1)}} = \frac{T_{H_(1)} - T_{C_(1)}}{T_{H_(1)}}` ..... (1)

And, formula for maximum efficiency of Carnot refrigerator is as follows.

`\frac{W}{Q_{C_(2)}} = \frac{T_{H_(2)} - T_{C_(2)}}{T_{C_(2)}}` ...... (2)

Now, equating both equations (1) and (2) as follows.

`Q_{C_(2)} \frac{T_{H_(2)} - T_{C_(2)}}{T_{C_(2)}}` =
`Q_{H_(1)} \frac{T_{H_(1)} - T_{C_(1)}}{T_{H_(1)}}`

`\gamma = \frac{Q_{C_(2)}}{Q_{H_(1)}}`

=
`\frac{T_{C_(2)}}{T_{H_(1)}} (\frac{T_{H_(1)} - T_{C_(1)}}{T_{H_(2)} - T_{C_(2)}})`

=
`(250)/(600) (((600 - 300)K)/(300 K - 250 K))`

= 2.5

Thus, we can conclude that the ratio of heat extracted by the refrigerator ("cooling load") to the heat delivered to the engine ("heating load") is 2.5.