Question

If a Carnot engine operating between two heat reservoir at 227°C absorbs 1000 calories from the 227°C reservoir per cycle, how much heat is discharged into the 27°C reservoir and how much work of done per cycle? What is the efficiency of the cycle?

Answer 1

Given 

T_h = 227°C or 500 k

T_c = 27°C or 300 k

Q_h = 1000 cal = heat absorb at 227°c

Q_c = heat rejected to 27°c reservoir

As we know , 

Efficiency of engine n = \(\frac{T_h-T_c}{T_h}\) = \(\frac{Q_h-Q_c}{Q_h}\)

∴ n = \(\frac{500-300}{500}\) 

= \(\frac{2}{5}\) 

= 0.4

∴ \(\frac{Q_h-Q_c}{Q_h}= 0.4\)

Q_h - Q_c = 0.4 x Q_h

Q_c = Q_h (1- 0.4)

= 0.6 x 1000 cal

Q_c = 600 cal

Hence 

heat dischaned into the 27°C reservoir = 600 cal

work done per cycle 

= Q_h - Q_c

= 1000 - 600

= 400 cal.