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A Carnot engine of efficiency 43% operates with a cold reservoir at 28°C and exhausts 1210...

A Carnot engine of efficiency 43% operates with a cold reservoir at 28°C and exhausts 1210 J of heat each cycle. What is the entropy change for the hot reservoir?

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Answer #1

= efficiency = 43% = 0.43

Tc = temperature of cold reservoir = 28 oC = 28 + 273 = 301 K

Th = temperature of hot reservoir = ?

efficiency is given as

= 1 - (Tc /Th )

0.43 = 1 - (301 /Th )

Th = 528.07 K

Qc = heat exhausted = 1210 J

Qh = heat input = ?

W = work done

efficiency is given as

= W/Qh

W = 0.43 Qh

we know that

W = Qh - Qc

0.43 Qh = Qh - 1210

Qh = 2122.81 J

entropy change is given as

S = Qh (Th - Tc)/(Th Tc)

S = (2122.81) (528.07 - 301) /(528.07 x 301)

S = 3.03

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