A refrigerator has to transfer an average of 263 J - s of heat from temperature -10- C to 25- C- What will be the average power consumed by the refrigerator- Assume no power loss in the processA- 15 WB- 35 WC -394 -5 WD- 920-5 W

The correct option is B 35 WGiven Q_2=263 J/s T_2= 10^∘C = 10+273=263 K T_1=25^∘C = 25+273=298 K β=Q_2/W=T_2/T_1 T_2 Work done per sec, W=Q_2(T_1 T_2)/T_2 ...

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Byju's Answer

Standard VII

Physics

Conduction

A refrigerato...

Question

A refrigerator has to transfer an average of $263 J / s$ of heat from temperature $- 10^{\circ}$ C to $25^{\circ}$ C. What will be the average power consumed by the refrigerator? (Assume no power loss in the process)

15 W

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35 W

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394.5 W

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920.5 W

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Solution

The correct option is B $35 W$
Given
$Q_{2} = 263 J / s$
$T_{2} = - 10^{\circ}$ C = $- 10 + 273 = 263$ K
$T_{1} = 25^{\circ}$ C = $25 + 273 = 298$ K

$β = \frac{Q_{2}}{W} = \frac{T_{2}}{T_{1} - T_{2}}$
Work done per sec, $W = \frac{Q_{2} (T_{1} - T_{2})}{T_{2}}$
$W = \frac{263 (298 - 263)}{263}$
$W = 35 W$

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A refrigerator has to transfer an average of 263 J/s of heat from temperature −10 ∘C to 25 ∘C. What will be the average power consumed by the refrigerator? (Assume no power loss in the process)

The correct option is B 35 WGiven Q2=263 J/s T2=−10∘C = −10+273=263 K T1=25∘C = 25+273=298 K β=Q2W=T2T1−T2 Work done per sec, W=Q2(T1−T2)T2 W=263 (298−263)263 W=35 W

A refrigerator has to transfer an average of $263 J / s$ of heat from temperature $- 10^{\circ}$ C to $25^{\circ}$ C. What will be the average power consumed by the refrigerator? (Assume no power loss in the process)