Given,
The efficiency of the engine, η=25%=0.25
The temperature of the cold reservoir, T₂=300 K
The efficiency of a heat engine is given by,
[tex]\eta=1-\frac{T_2}{T_1}[/tex]Where T₁ is the temperature of the hot reservoir
On rearranging the above equation,
[tex]T_1=\frac{T_2}{1-\eta}_{}[/tex]On substituting the known values in the above equation,
[tex]T_1=\frac{300}{1-0.25}=400\text{ K}[/tex]Therefore the temperature of the hot reservoir is 400 K
