A completely reversible heat pump produces heat at a rate of 100 kW to warm a house maintained at 25℃. The exterior air, which is at 0℃, serves as the source. Calculate the rate of entropy change of the two reservoirs and determine if this heat pump satisfies the second law according to the increase of entropy principle.

Question

Kathryn Wall

Engineering

14 February, 22:45

A completely reversible heat pump produces heat at a rate of 100 kW to warm a house maintained at 25℃. The exterior air, which is at 0℃, serves as the source. Calculate the rate of entropy change of the two reservoirs and determine if this heat pump satisfies the second law according to the increase of entropy principle.

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Hess · Answer 1

1) 0. 03 kW/K

2) the value is grater than zero so it satisfies the second law of thermodynamic (states that rate of entropy change must be equal to or greater than zero).

Explanation:

Rate of entropy change S = dQ/dT

= Q (1/T1 - 1/T2)

T2 = 25°C = 298 K

T1 = 0°C = 273 K

Q = 100 kW

S = 100 (1/273 - 1/298)

S = 100 (0.0003) = 0. 03 kW/K