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A refrigerator has to transfer an average of 263 J of heat per second from temperature −10°C to 25°C. Calculate the average ideal reversible cycle and no other losses.

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Best answer

Given : 

T2 = 25 + 273 = 298 K

T1 = −10 + 273 = 263 K

From relation,

\(\frac{Q_2}{Q_1}=\frac{T_2}{T_1}\) 

Or,  Q1\(\frac{T_2}{T_1}\) × Q2

\(\frac{298}{263}\times263\) 

= 298 Js-1

∴ Average power consumed = Q1 − Q2

= (298 – 263) Js-1

= 35 W.

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