Question

A piece of equipment functions such that a hot reservoir at 605 K transfers 1150 J of heat irreversibly to a cold reservoir at a temperature of 296 K.

1)

What is the change in entropy of the hot reservoir?

ΔS_H = J/K

2)

What is the change in entropy of the cold reservoir?

ΔS_c = J/K

3)

What is the change in entropy of the whole universe due to this process?

ΔS = J/K

Answer 1

I'll call Q₂ the transfered heat. Despite the fact the transformation is irreversible, I'll use for reversible process.

OBS: The problem is wrong: it stipulates an irreversible transformation, but asks for changes in entropy as if it was a reversible process (for both reservoirs). I'll comment on this in details at the end of the solution (see point 3).

Generally, , for a reversible process. If irreversible, .

1) For the hot reservoir: Q<0 -->

2) For the cold reservoir: Q>0 -->

3) Not too much. Negligible.

To give a more detailed answer at this point I should know if the system undergoing this tranformation is isolated from the environment or not. I mean, the reservoirs.

If the system is isolated from the external environment, there will be no change in the "whole universe" entropy due to the process this system undergoes.

If the system is not isolated, then we have 2 isothermal irreversible processes: one for each reservoir.

If the transformations are reversible, we have for the environment, for each reservoir:

Thus, the total sum is 0.

If the transformations are irreversible there is a so-called "lost work" (work it could have been obtained, but did not).

Generally: (if you know the parameters you may calculate it; usually has small values, around few J/K).

For each reservoir we have:

(the last term: the entropy generated due to irreversibility), and .