The temperature inside the refrigerator is 2.0 C. It stands in a kitchen whose t

Question

The temperature inside                 the refrigerator is 2.0 C. It stands in a kitchen whose temperature is 22 C. After a period of an hour, 95 kJ of heat are transferred from the interior of the                 refrigerator to the kitchen.

As a result of this heat transfer, what is the entropy change of (i) the contents of the refrigerator, and (ii) the kitchen? (Assume                 that the heat transfer is reversible.)

State the principle of entropy increase. Are your answers to the above compatible with this principle? If not,                 suggest an explanation (one or two sentences).             

Explanation / Answer

Change in entropy delta S = delta Q/T if the process is reversible.As the process is reversible. delta S for contents of refrigerator = delta Q /T = -95 kJ/(2+273) K = -345.455 J/K -ve sign because heat is lost by the refrigerator contents.

similarly delta S for kitchen = delta Q/T = 95 kJ/(22+273)K = 322.034 J/k =ve sign because heat is received.

net change in entropy = delta S of kitchen + delta S of contents = 322.034 J/k -345.455 J/K

= -23.421 J/K

increase of entropy principle states that the process is spontaneous only if the nett chenge in entropy is greater than or equal to 0.

It appears that our answer is not feasible with the principle, but actually net change in entropy we calculated is not at all the actual nett change.There is a fluid or refrigerant which cools the contents and lets out heat to kitchen.We must also take into account the change in entropy of that refrigerant.The work done on the refrigerant enables the operation.

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