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You have a hot reservoir held at 30°C and a cold reservoir held at 0°C. If you move 400 J of heat from the hot reservoir to the cold reservoir, what is the total change in entropy?

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2 Answers

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Answer:

-400/303 = -1.32

400/273 = 1.47

-1.32 + 1.47 =

.15 J/K

Step-by-step explanation:

a p e x

2 votes

Answer:

Step-by-step explanation:

Given the following :

Temperature (Th) of hot reservoir = 30°C (30 +273) = 303K

Temperature (Tc) of cold reservoir = 0°C (273K)

Quantity (Q) of heat transferred from hot reservoir = 400 J

Total change in entropy (ΔStotal) :

ΔStotal = ΔShot + ΔScold

ΔS = Q/T

Hot reservoir is losing 400J of heat ;

Q = - 400 J

ΔShot = Q/Th

ΔShot = - 400/303

ΔShot = - 1.32 J/K

ΔScold = Q/T

ΔScold = 400 / 273

ΔScold = 1.47 J/K

ΔStotal = ΔShot + ΔScold

ΔStotal = - 1.32 + 1.47

ΔS total = 0.15 J/K

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