A reversed Carnot cycle refrigerator maintains a temperature of -5-C- The ambient air temperature is 35-C- The heat gained by the refrigerator at a continuous rate is 2-5 kJ-s- The power in watts required to pump this heat out continuously is

T_2= 5^∘C=( 5+273)K=268 K T_1=35^∘C=(35+273)K=308 K Q_2=2.5 kJ/s =2.5 kW=2500 W (COP)_R = T_2T_1 T_2 =268308 268 = 6.7 also (COP)_r=Q_2W ∴ 6.7 = 2500W ...

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Standard XII

Physics

Heat Capacity

A reversed Ca...

Question

A reversed Carnot cycle refrigerator maintains a temperature of $- 5^{\circ} C$ . The ambient air temperature is $35^{\circ} C$ . The heat gained by the refrigerator at a continuous rate is 2.5 kJ/s. The power (in watts) required to pump this heat out continuously is

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Solution

$T_{2} = - 5^{\circ} C = (- 5 + 273) K = 268 K$

$T_{1} = 35^{\circ} C = (35 + 273) K = 308 K$

$Q_{2} = 2.5 k J / s$

$= 2.5 k W = 2500 W$

$(C O P)_{R} = \frac{T_{2}}{T_{1} - T_{2}}$

$= \frac{268}{308 - 268}$

= 6.7
also $(C O P)_{r} = \frac{Q_{2}}{W}$

$∴ 6.7 = \frac{2500}{W}$
or
$W = \frac{2500}{6.7} = 373.13 W$

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A reversed Carnot cycle refrigerator maintains a temperature of −5∘C. The ambient air temperature is 35∘C. The heat gained by the refrigerator at a continuous rate is 2.5 kJ/s. The power (in watts) required to pump this heat out continuously is

T2=−5∘C=(−5+273)K=268K T1=35∘C=(35+273)K=308K Q2=2.5kJ/s =2.5kW=2500W (COP)R=T2T1−T2 =268308−268 = 6.7 also (COP)r=Q2W ∴6.7=2500W or W=25006.7=373.13W

A reversed Carnot cycle refrigerator maintains a temperature of $- 5^{\circ} C$ . The ambient air temperature is $35^{\circ} C$ . The heat gained by the refrigerator at a continuous rate is 2.5 kJ/s. The power (in watts) required to pump this heat out continuously is