Second Law of Thermodynamics - Refrigerators

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Homework Statement



Suppose 10000 J is transferred between a large thermal reservoir at a temperature of
300 K to another large thermal reservoir at 500 K. What is the net change in entropy
of the system? Write an expression for the factor by which the multiplicity of the
system changes.

For this system, what is the minimum amount of work required to transfer the energy
between the two reservoirs, and under what conditions will this calculation be valid?

Homework Equations



COP = Qc/W = Qc/(Qh-Qc) = Tc/(Th-Tc)

DeltaS = -(Qc/Tc) + (Qh/Th)

W = Qc/W

Qh = Qc + W


The Attempt at a Solution



COP = k = 300/(500-300) = 1.5
Qc = 10000
W= 10000/1.5 = 6666.6
Qh = 10000 + 6666.6 = 16666.6

Therefore

DeltaS = -(10000/300) + (16666.6/500)
= -33.3333 + 33.3332

This doesn't seem right to me but I don't know why.

Also this would give me a multiplicity increase by a factor of e^(0/k) which is 1. So that surely can't be right.

I apologise if this is really simple and I'm making stupid mistakes.
 
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This is not a refrigerator problem. This is just a direct transfer of heat from a hot reservoir to a cold reservoir. The OP, for some reason, got the idea that it is a reservoir problem. No.