Question

Two reversible heat engines are operated with same heat input and heat output. One of the engine is operated between the temperature limits of \(T_2\) and 200 K & another is between 800 K and \(T_2\). Then the value of \(T_2\) is

• A. \(100 \text{ K} \)
• B. \(200 \text{ K} \)
• C. \(400 \text{ K} \)
• D. \(500 \text{ K} \)

Hint:

When two reversible engines have the same efficiency (implied here by equal heat inputs and outputs), the intermediate temperature $T_2$ is always the geometric mean of the extreme temperatures: $T_2 = \sqrt{T_{\text{max}} \cdot T_{\text{min}}}$. Here, $T_2 = \sqrt{800 \times 200} = \sqrt{160000} = 400 \text{ K}$.

Answer 1

The Correct Option is C

Concept: For a completely reversible heat engine operating on a Carnot cycle, the thermal efficiency ($\eta$) depends strictly on the absolute temperature limits of the thermal reservoirs it interacts with: \[ \eta = 1 - \frac{T_L}{T_H} \] Where $T_L$ is the lower sink temperature and $T_H$ is the upper source temperature. The efficiency is also fundamentally defined as: \[ \eta = \frac{W_{\text{net}}}{Q_{\text{in}}} = 1 - \frac{Q_{\text{out}}}{Q_{\text{in}}} \] The statement tells us that both engines have the same heat input ($Q_{\text{in}}$) and the same heat output ($Q_{\text{out}}$). This implies that their thermal efficiencies are identical ($\eta_1 = \eta_2$). Consequently, for reversible engines, the ratio of sink temperature to source temperature must also be equal across both units.

Step 1: Set up equations for both engines using temperature values.
Let's analyze the configuration of both thermodynamic units:

• Engine 1: Fits between source temperature $T_2$ and sink temperature $200 \text{ K}$. \[ \eta_1 = 1 - \frac{200}{T_2} \]

• Engine 2: Fits between source temperature $800 \text{ K}$ and sink temperature $T_2$. \[ \eta_2 = 1 - \frac{T_2}{800} \]

Step 2: Equate efficiencies to establish the algebraic relationship.
Since $Q_{\text{in}, 1} = Q_{\text{in}, 2}$ and $Q_{\text{out}, 1} = Q_{\text{out}, 2}$, we set $\eta_1 = \eta_2$: \[ 1 - \frac{200}{T_2} = 1 - \frac{T_2}{800} \] Subtracting 1 from both sides gives: \[ -\frac{200}{T_2} = -\frac{T_2}{800} \] Eliminating the negative signs from both sides yields: \[ \frac{200}{T_2} = \frac{T_2}{800} \]

Step 3: Cross-multiply and solve for intermediate temperature \(T_2\).
Cross-multiplying across the fractions: \[ T_2^2 = 800 \times 200 \] \[ T_2^2 = 160000 \] Taking the positive square root on both sides since thermodynamic temperatures must remain positive: \[ T_2 = \sqrt{160000} = 400 \text{ K} \] Thus, the intermediate temperature $T_2$ is equal to 400 K, which corresponds to Option (C).