First slide

Heat engine

Question

A refrigerator works between $4^{0} C and 30^{0} C$ . It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is

Moderate

A

2.365 W
B

23.65 W
C

236.5 W
D

2365 W

Solution

$β = \frac{Q_{2}}{W} = \frac{T_{2}}{T_{1} - T_{2}}$
(where $Q_{2}$ is heat removed)
$\Rightarrow \frac{600 \times 4.2}{W} = \frac{277}{303 - 277}$
$\Rightarrow W = 236.5 joule$
$\Rightarrow Power = \frac{W}{t} = \frac{236.5 joule}{1 \sec} = 236.5 W$

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