Answer:
[tex]T_H = 1335\ J[/tex]
Explanation:
Given,
Work done by engine = 19900 J
rejected energy of heat = 5300 J
temperature = 285 K
efficiency of the engine
[tex]\eta = \dfrac{W}{Q_H}[/tex]
where
[tex]W = Q_H - Q_C[/tex]
[tex]19900 = Q_H - 5300[/tex]
[tex]Q_H = 25200\ J[/tex]
efficiency in terms of temperature
[tex]\eta = \dfrac{T_H-T_C}{T_H}[/tex]
from the above two equation
[tex]\dfrac{Q_H-Q_C}{Q_H}= \dfrac{T_H-T_C}{T_H}[/tex]
[tex]1-\dfrac{Q_C}{Q_H}= 1- \dfrac{T_C}{T_H}[/tex]
[tex]\dfrac{Q_C}{Q_H}= \dfrac{T_C}{T_H}[/tex]
[tex]T_H=T_C\ \dfrac{Q_H}{Q_C}[/tex]
[tex]T_H=285\times \dfrac{25200}{5300}[/tex]
[tex]T_H = 1335\ J[/tex]
