Answer:
the temperature of the intermediate reservoir is 624.5 K
Explanation:
Given the data in the question
The two Carnot heat engines are operating in series;
[ T[tex]_H[/tex] ]
↓
((1)) ⇒ W[tex]_{out[/tex]
↓
[ T[tex]_M[/tex] ]
↓
((2)) ⇒ W[tex]_{out[/tex]
[ T[tex]_L[/tex] ]
The maximum possible efficiency for any heat engine is the Carnot efficiency;
η[tex]_{rev[/tex] = 1 - [tex]\frac{T_L}{T_H}[/tex]
the thermal efficiencies if both engines are the same will be;
η[tex]_A[/tex] = η[tex]_B[/tex]
1 - [tex]\frac{T_M}{T_H}[/tex] = 1 - [tex]\frac{T_L}{T_M}[/tex]
1 - 1 - [tex]\frac{T_M}{T_H}[/tex] = - [tex]\frac{T_L}{T_M}[/tex]
- [tex]\frac{T_M}{T_H}[/tex] = - [tex]\frac{T_L}{T_M}[/tex]
[tex]\frac{T_M}{T_H}[/tex] = [tex]\frac{T_L}{T_M}[/tex]
T[tex]_M[/tex]² = T[tex]_L[/tex] × T[tex]_H[/tex]
T[tex]_M[/tex] = √(T[tex]_L[/tex] × T[tex]_H[/tex])
source temperature of the first engine T[tex]_H[/tex] = 1300 K
sink temperature of the second engine T[tex]_L[/tex] = 300 K
we substitute
T[tex]_M[/tex] = √(300 × 1300)
T[tex]_M[/tex] = √390000
T[tex]_M[/tex] = 624.4998 K ≈ 624.5 K
Therefore, the temperature of the intermediate reservoir is 624.5 K
