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Internal Energy

An ideal refr...

Question

An ideal refrigerator is operating between $37^{\circ} C$ and $- 13^{\circ} C$ . The power required per ton of refrigerator is

A

1.2 kW

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B

0.9 kW

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C

0.4 kW

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D

0.67 kW

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Solution

The correct option is D 0.67 kW

$T_{H} = 273 + 37 = 310 K$

$T_{L} = 273 - 13 = 260 K$

COP $= \frac{R C}{P_{in}} = \frac{T_{L}}{T_{H} - T_{L}}$

$\Rightarrow \frac{3.5}{P_{in}} = \frac{260}{310 - 260}$

$\Rightarrow P_{in} = \frac{3.5 \times 50}{260} = 0.673 kW$

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