Final Answer:
The value of T at which the efficiencies of the two engines are equal is √T₁T₂. Option C is correct.
Step-by-step explanation:
In a series combination of Carnot engines, the overall efficiency is given by the product of the individual efficiencies. The efficiency of a Carnot engine is given by 1 - (T₂/T₁), where T₁ is the temperature at which it receives heat and T₂ is the temperature at which it rejects heat.
So, the efficiency of engine A is 1 - (T/T₁), and the efficiency of engine B is 1 - (T₂/T). To find the value of T at which the efficiencies of the two engines are equal, we can equate the two expressions:
1 - (T/T₁) = 1 - (T₂/T)
Canceling out the 1's, we have:
T/T₁ = T₂/T
Cross-multiplying, we get:
T * T = T₁ * T₂
Simplifying, we have:
T = √(T₁ * T₂)
Therefore, the value of T at which the efficiencies of the two engines are equal is √T₁T₂.
So, the correct answer is (c) √T₁T₂.
Complete question:
Two Carnot engines A and B are operated in series. The engine A receives heat from the source at temperature T₁ and rejects the heat to the sink at temperature T. The second engine B receives the heat at temperature T and rejects to its sink at temperature T₂. For what value of T the efficiencies of the two engines are equal.
- (a) (T₁- T₂) / 2
- (b) T₁T₂
- (c) √T₁T₂
- (d) (T₁+T₂) / 2