A refrigerator works between $0^{o} C$ and $27^{o} C$ . Heat is to be removed from refrigerated space at the rate of $50 k c a l / m i n$ , the power of the motor of the refrigerator is:

0.346 k W

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3.46 k W

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34.6 k W

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346 k W

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Solution

The correct option is A $0.346 k W$
$\frac{T_{2}}{T_{1} - T_{2}} = \frac{Q_{2}}{W}$

$\frac{273}{300 - 273} = \frac{50, 000}{W}$

$W = \frac{27 \times 50, 000}{273} c a l / m i n$
$p = \frac{W}{t} = \frac{4.2 \times 27 \times 50, 000}{60 \times 273} J / s$
$= 346 w a t t = 0.346 k W$

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A refrigerator works between 0oC and 27oC. Heat is to be removed from refrigerated space at the rate of 50 kcal/min, the power of the motor of the refrigerator is:

The correct option is A 0.346 kWT2T1−T2=Q2W273300−273=50,000WW=27×50,000273cal/minp=Wt=4.2×27×50,00060×273J/s=346 watt=0.346 kW

A refrigerator works between $0^{o} C$ and $27^{o} C$ . Heat is to be removed from refrigerated space at the rate of $50 k c a l / m i n$ , the power of the motor of the refrigerator is:

The correct option is A $0.346 k W$
$\frac{T_{2}}{T_{1} - T_{2}} = \frac{Q_{2}}{W}$

$\frac{273}{300 - 273} = \frac{50, 000}{W}$

$W = \frac{27 \times 50, 000}{273} c a l / m i n$
$p = \frac{W}{t} = \frac{4.2 \times 27 \times 50, 000}{60 \times 273} J / s$
$= 346 w a t t = 0.346 k W$