Answer:
the temperature of the intermediate reservoir is 624.5 K
Step-by-step explanation:
Given the data in the question
The two Carnot heat engines are operating in series;
[ T
]
↓
((1)) ⇒ W
![_{out](https://img.qammunity.org/2022/formulas/physics/college/weobztzl30ka3z2eduswp94jn6u21kclmh.png)
↓
[ T
]
↓
((2)) ⇒ W
![_{out](https://img.qammunity.org/2022/formulas/physics/college/weobztzl30ka3z2eduswp94jn6u21kclmh.png)
[ T
]
The maximum possible efficiency for any heat engine is the Carnot efficiency;
η
= 1 -
![(T_L)/(T_H)](https://img.qammunity.org/2022/formulas/physics/college/7lelawbawp16oyt4jftzso976o01veflkd.png)
the thermal efficiencies if both engines are the same will be;
η
= η
![_B](https://img.qammunity.org/2022/formulas/chemistry/college/zz97z5j3gqawzzykn7473ip0c6uqm7agl9.png)
1 -
= 1 -
![(T_L)/(T_M)](https://img.qammunity.org/2022/formulas/physics/college/5k1ysprx0ff25mpw3cxmvcx0jxyyi7ahet.png)
1 - 1 -
= -
![(T_L)/(T_M)](https://img.qammunity.org/2022/formulas/physics/college/5k1ysprx0ff25mpw3cxmvcx0jxyyi7ahet.png)
-
= -
![(T_L)/(T_M)](https://img.qammunity.org/2022/formulas/physics/college/5k1ysprx0ff25mpw3cxmvcx0jxyyi7ahet.png)
=
![(T_L)/(T_M)](https://img.qammunity.org/2022/formulas/physics/college/5k1ysprx0ff25mpw3cxmvcx0jxyyi7ahet.png)
T
² = T
× T
![_H](https://img.qammunity.org/2022/formulas/physics/college/v5wmgwthx60rrj7k9qdege9xq4wxhporpn.png)
T
= √(T
× T
)
source temperature of the first engine T
= 1300 K
sink temperature of the second engine T
= 300 K
we substitute
T
= √(300 × 1300)
T
= √390000
T
= 624.4998 K ≈ 624.5 K
Therefore, the temperature of the intermediate reservoir is 624.5 K