A thermal energy reservoir at 1200 K supplies 500 kJ of heat to a reversible heat engine E1. Engine E1 rejects heat Q2 to a reversible heat engine E2 at temperature T2. Engine E2 rejects heat to a thermal energy reservoir at temperature 300 K. If the efficiency of both the engines is the same then T1 : T2 : T3 is: 

Concept:

The intermediate temperature (T₂) for two reversible engines operating in a series combination between higher temperature (T₁) and lower temperature (T₃) if the efficiency of both the engines are same then is given by

⇒ \(T_2 = \sqrt{T_1T_3}\;\).      ...(1)

Calculation:

Given:

T1 = 1200 K T3 = 300 K

E1 and E2 engines are in a series combination and the efficiency of both engines is the same.

As the efficiencies for both reversible engines are the same we have to put the values of higher and lower temperatures in equation (1) to calculate the intermediate temperature.

\(T_2 = \sqrt{T_1T_3}\;\)

⇒ \(T_2 = \sqrt{1200\times 300} = 600K\;\)

So T₁ : T₂ : T₃ can be written as 

⇒ 1200: 600: 300

On dividing the whole ratio by 600 then

We get

⇒ T1 : T2 : T3 = 2 : 1 : 0.5