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

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.

 $Q_{C_{2}} \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}}$ = $Q_{H_{1}} \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}}$        

        $\gamma = \frac{Q_{C_{2}}}{Q_{H_{1}}}$

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

                    = $\frac{250}{600} (\frac{(600 - 300)K}{300 K - 250 K})$

                    = 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.