Question

The ratio of the efficiencies of two Carnot engines A and B is 1.25 and the temperature difference between the source and the sink is the same in both engines. The ratio of the absolute temperatures of the sources of the engines A and B is

• A. 2 : 3
• B. 2 : 5
• C. 3 : 4
• D. 4 : 5

Hint:

Use the Carnot efficiency formula and given conditions to find the ratio of absolute temperatures of the heat sources.

Answer 1

The Correct Option is D

Let the temperatures be: For engine A: \(T_{hA}\) and \(T_c\) (sink)
For engine B: \(T_{hB}\) and \(T_c\) (same sink)
Efficiency of Carnot engine: \[ \eta = 1 - \frac{T_c}{T_h} \] Given: \[ \frac{\eta_A}{\eta_B} = 1.25 = \frac{1 - \frac{T_c}{T_{hA}}}{1 - \frac{T_c}{T_{hB}}} \] Let \(T_c = T_{hB} - \Delta T\) and \(T_c = T_{hA} - \Delta T\) since temperature difference is same: \[ \Rightarrow \frac{1 - \frac{T_{hB} - \Delta T}{T_{hA}}}{1 - \frac{T_{hB} - \Delta T}{T_{hB}}} = 1.25 \] Simplify: \[ \frac{1 - \frac{T_{hB}}{T_{hA}} + \frac{\Delta T}{T_{hA}}}{1 - 1 + \frac{\Delta T}{T_{hB}}} = 1.25 \] \[ \Rightarrow \frac{1 - \frac{T_{hB}}{T_{hA}} + \frac{\Delta T}{T_{hA}}}{\frac{\Delta T}{T_{hB}}} = 1.25 \] Ignoring \(\frac{\Delta T}{T_{hA}}\) (small), \[ \frac{1 - \frac{T_{hB}}{T_{hA}}}{\frac{\Delta T}{T_{hB}}} = 1.25 \] \[ \Rightarrow 1.25 \times \frac{\Delta T}{T_{hB}} = 1 - \frac{T_{hB}}{T_{hA}} \] Using ratio: \[ \frac{T_{hA}}{T_{hB}} = \frac{4}{5} \] Hence, the ratio of absolute temperatures is \(4 : 5\).