Answer to Question #273352 in Physics for Faru

Question

an ideal heat engine has an efficiency of 25% 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

The efficiency is

$$\eta=1-\frac{T_c}{T_h},\\\space\\ T_h=\frac{T_c}{1-\eta}=397\text{ K}$$

for 25%. For 35%, the following changes can be made:

$$T_h=\frac{T_c}{1-\eta}=458\text{ K},\\\space\\ T_c=(1-\eta)T_h=258\text{ K}$$

to the temperatures of the hot and cold receiver respectively.