Answer:
The temperature at which observed heat is 400 K
Explanation:
Given data:
rejection reservoir temperature at exit [tex]T_{L}[/tex] is 300 k
the efficiency of a engine is η = 25%
we know that efficiency of Carnot is given as
[tex]\eta = (1-\frac{T_{L}}{T_{H}})*100[/tex]
Putting all value to obtained temperature at which observed heat
[tex]0.25 = (1-\frac{300}{T_{H}})[/tex]
[tex]T_{H}[/tex] = 400 K
