One reversible heat engine operates between 1600 K and T2 K and another reversible heat engine operates between T2 K and 400 K. If both the engines have the same heat input and output, then temperature T2 is equal to

Short cut methods
√(1600•400)=40•20=800K

If there is two reversible heat engine or we can call it tapping cycle then there are two cases for determining the mid temperature...
1. If work done of both the heat engines are same
Then mid temperature(T2) is average of source temperature (1600K) and last heat engine's sink temperature (400K)
2. If efficiencies of both the heat engines are same then mid temperature (T2) is the geometric mean of source temperature (1600K) and last heat engine's sink temperature (400K).

Otherwise we can calculate combined
efficiency...

(1- eff. Of combined heat engine) = (1- eff. Of 1st engine ) (1-eff. Of 2nd engine)

We know all reversible engine have same efficiency when the same input so,

(1600-T2)/1600=(T2-400)/T2

Solve this,
We get
T2^2=640000
Then,
T2=√(640000)
T2= 800K

You know the Carnot rule that said all reversible engine have same efficiency so by using trail and error way easy to find the answer

One reversible heat engine operates between 1600 K and T2 K and another reversible heat engine operates between T2 K and 400 K. If both the engines have the same heat input and output, then temperature T2 is equal to
A. 800 K

B. 1000 K

C. 1200 K

D. 1400 K

Please reply must

Please give explains how 800 k what is the rule and law