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A copper block receives heat from two different sources: 5 kW from a source at 1500 K and 3 kW from a source at 1000 K. It loses heat to atmosphere at 300 K. Assuming the block to be at steady state, determine (a) the net rate of heat transfer in kW; (b) the rate of entropy generation in the system's universe

User David Boho
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Answer:

a) Zero

b) the rate of entropy generation in the system's universe = ds/dt = 0.2603 KW/K

Step-by-step explanation:

a) In steady state

Net rate of Heat transfer = net rate of heat gain - net rate of heat lost

Hence, the rate of heat transfer = 0

b) In steady state, entropy generated

ds/dt = - [ Qgain/Th1 + Qgain/Th2 - Qlost/300 K]

Substituting the given values, we get –

ds/dt = -[5/1500 + 3/1000 – (5+3)/300]

ds/dt = - [0.0033 + 0.003 -0.2666]

ds/dt = 0.2603 KW/K

User Slecorne
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