Answer:
Given:
high temperature reservoir [tex]T_{H} =1000k[/tex]
low temperature reservoir [tex]T_{L} =400k[/tex]
thermal efficiency [tex]n_{1}= n_{2}[/tex]
The engines are said to operate on Carnot cycle which is totally reversible.
To find the intermediate temperature between the two engines, The thermal efficiency of the first heat engine can be defined as
[tex]n_{1} =1-\frac{T}{T_{H} }[/tex]
The thermal efficiency of second heat engine can be written as
[tex]n_{2} =1-\frac{T_{L} }{T}[/tex]
The temperature of intermediate reservoir can be defined as
[tex]1-\frac{T}{T_{H} } =1-\frac{T_{L} }{T} \\T^2=T_{L} T_{H} \\T=\sqrt{T_{L} T_{H} }\\T=\sqrt{400*1000} =632k[/tex]