Answer:
Assuming that 4000 J of heat transfer occurs from it the change in entropy will be=2.33j/k
Step-by-step explanation:
Given data
Th=619 K
Tc= 455K
We are going to assume that 4000 J of heat transfer occurs from it, since it was not specified in the question.
we know that the change in entropy is given as
ΔStotal=ΔShot+ΔScold .
ΔShot=−Q/Th=−4000J/619K=−6.46J/K
For the cold reservoir,
ΔScold=Q/Tc=4000J/455K=8.79J/K
therefore the total is
ΔStotal=ΔShot+ΔScold
=(−6.46+8.79)J/K
=2.33j/k