Question

Two Carnot heat engines (E1 and E2) are operating in series as shown in the figure. Engine E1 receives heat from a reservoir at \(T_H = 1600 \, {K}\) and does work \(W_1\). Engine E2 receives heat from an intermediate reservoir at \(T\), does work \(W_2\), and rejects heat to a reservoir at \(T_L = 400 \, {K}\). Both the engines have identical thermal efficiencies. The temperature \(T\) (in K) of the intermediate reservoir is ........ (answer in integer). 

Hint:

For Carnot engines, when efficiencies are equal, equate the efficiency formulas and solve for the unknown temperature.

Answer 1

Correct Answer: 800

Step 1: Using the efficiency formula for a Carnot engine: The efficiency \( \eta \) of a Carnot engine is given by: \[ \eta = 1 - \frac{T_L}{T_H} \] where \( T_H \) is the temperature of the hot reservoir and \( T_L \) is the temperature of the cold reservoir. For engine 1: \[ \eta_1 = 1 - \frac{T}{1600} \] where \( T \) is the temperature of the intermediate reservoir. For engine 2: \[ \eta_2 = 1 - \frac{400}{T} \] Since the engines have identical efficiencies, we set \( \eta_1 = \eta_2 \): \[ 1 - \frac{T}{1600} = 1 - \frac{400}{T} \] Step 2: Solving for \( T \): \[ \frac{T}{1600} = \frac{400}{T} \] Cross-multiply to solve for \( T \): \[ T^2 = 1600 \times 400 \] \[ T = \sqrt{1600 \times 400} = \sqrt{640000} = 800 \, {K} \]