Question

A Carnot engine whose high-temperature reservoir is at 464 K has an efficiency of 25.0%. By how much should the temperature of the low-temperature reservoir be changed to increase the efficiency to 42.0%?

Answer 1

Answer

given,

high temperature reservoir (T_c)= 464 K

efficiency  of reservoir (ε)= 25 %

temperature to decrease = ?

increase in efficiency = 42 %

now, using equation

 $\epsilon = 1 - \dfrac{T_C}{T_H}$

 $0.25 = 1 - \dfrac{T_C}{464}$

 $T_C= (1 - 0.25) \times 464$

 $T_C= 0.75 \times 464$

      T_C = 348 K

now,

if the efficiency is equal to 42$ = 0.42

 $T_C= (1 - 0.42) \times 464$

 $T_C= 0.58 \times 464$

 $T_C= 269.12\ K$