Question

Suppose a heat source generates heat at a rate of 87.0 w (1 w = 1 j/s). how much entropy does this produce per hour in the surroundings at 27.4 °c? assume the heat transfer is reversible.

Answer 1

For a reversible process, the second law says that:
dS = dq/T
So
dS/dt = (1/T)*dq/dt
We are given that dq/dt = 87.0 J/s, and that T = 27.4°C which is equal to 300.55 K, so:
dS/dt = (1/300.55K)*(87.0 J/s) = 0.289 J/(K*s) which is equal to 1040.40 joules per hour