Question

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?

Answer 1

= efficiency = 43% = 0.43

T_c = temperature of cold reservoir = 28 ^oC = 28 + 273 = 301 K

T_h = temperature of hot reservoir = ?

efficiency is given as

= 1 - (T_c /T_h )

0.43 = 1 - (301 /T_h )

T_h = 528.07 K

Q_c = heat exhausted = 1210 J

Q_h = heat input = ?

W = work done

efficiency is given as

= W/Q_h

W = 0.43 Q_h

we know that

W = Q_h - Q_c

0.43 Q_h = Q_h - 1210

Q_h = 2122.81 J

entropy change is given as

S = Q_h (T_h - T_c)/(T_h T_c)

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

S = 3.03