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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%?

Respuesta :

Answer

given,

high temperature reservoir (T_c)= 464 K

efficiency  of reservoir (ε)= 25 %

temperature to decrease = ?

increase in efficiency = 42 %

now, using equation

 [tex]\epsilon = 1 - \dfrac{T_C}{T_H}[/tex]

 [tex]0.25 = 1 - \dfrac{T_C}{464}[/tex]

 [tex]T_C= (1 - 0.25) \times 464 [/tex]

 [tex]T_C= 0.75 \times 464 [/tex]

      T_C = 348 K

now,

if the efficiency is equal to 42$ = 0.42

 [tex]T_C= (1 - 0.42) \times 464 [/tex]

 [tex]T_C= 0.58 \times 464 [/tex]

 [tex]T_C= 269.12\ K [/tex]

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