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Refrigerator – heat rejected per unit work
Concepts tested here
- refrigerator-heat-to-room
All Questions
2016 Phase II 1 question
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The temperature inside a refrigerator is $t_2$°C and the room temperature is $t_1$°C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be:
$Q_1/W = T_1/(T_1 - T_2)$.
Heat delivered to the room $= Q_1$
COP $\beta = \frac{t_2 + 273}{t_1 - t_2} = \frac{Q_2}{W} = \frac{Q_1 - W}{W} = \frac{Q_1}{W} - 1$
$\Rightarrow \frac{Q_1}{W} = 1 + \frac{t_2 + 273}{t_1 - t_2} = \frac{t_1 + 273}{t_1 - t_2}$
