Question

an ideal heat engine has an efficiency of 50% when its low temperature reserver is 25°C. it is desire to increase the efficiency to 35%. what change should be made in either (1) the temperature of the low reserver or (2) the temperature of the high reserver??

Answer 1

$$\eta=1-\frac{T_c}{T_h}\\0.5=1-\frac{25+273}{T_h}\\T_h=596\ K$$

1)

$$0.35=1-\frac{T_c}{596}\\T_h=387\ K$$

2)

$$0.35=1-\frac{273+25}{T_h}\\T_h=458\ K$$