(a) 287.9 K
The efficiency of the engine is given by
[tex]\eta = 1-\frac{T_C}{T_H}[/tex]
where
Tc is the temperature of the cold reservoir
Th is the temperature of the hot reservoir
We can rewrite the two temperatures as
[tex]T_c = T_h - 68.8 K[/tex]
(because differences in Kelvin are equal to differences in Celsius)
So we can rewrite the 1st equation as
[tex]\eta = 1-\frac{T_H - 68.8}{T_H} =1-1+\frac{68.8}{T_H}=\frac{68.8}{T_H}[/tex]
We also know the efficiency:
[tex]\eta = 23.9\% = 0.239[/tex]
So we can re-arrange the previous equation to find the temperature of the hot reservoir:
[tex]T_H = \frac{68.8}{0.239}=287.9 K[/tex]
(b) 219.1 K
The temperature of the cold reservoir can be found as:
[tex]T_C = T_H - 68.8 = 287.9 - 68.8=219.1 K[/tex]
