A reversible heat engine A (based on carnot cycle) absorbs heat from a reservoir at 1000 K and rejects heat to a reservoir at `T_(2)`. A second reversible engine B absorbs, the same amount of heat as rejected by the engine A, from the reservoir at `T_(2)` and rejects energy to a reservoir at 360 K. If the efficiencies of engines A and B are the same then the temperautre `T_(2)` is

Options

A. 680 K
B. 640 K
C. 600 K
D. None

Answer

To solve the problem, we need to find the temperature \( T_2 \) at which the second reversible heat engine B operates, given that both engines A and B have the same efficiency.

### Step-by-step Solution:

1. **Understand the Efficiency of Engine A:**
The efficiency \( \eta_A \) of engine A, which operates between a hot reservoir at temperature \( T_1 = 1000 \, K \) and a cold reservoir at temperature \( T_2 \), is given by the formula:
\[ \eta_A = 1 - \frac{T_2}{T_1} = 1 - \frac{T_2}{1000} \]

2. **Understand the Efficiency of Engine B:**
The efficiency \( \eta_B \) of engine B, which absorbs heat from the reservoir at temperature \( T_2 \) and rejects heat to a reservoir at temperature \( T_3 = 360 \, K \), is given by:
\[ \eta_B = 1 - \frac{T_3}{T_2} = 1 - \frac{360}{T_2} \]

3. **Set the Efficiencies Equal:**
Since the efficiencies of both engines are equal, we can set the equations for \( \eta_A \) and \( \eta_B \) equal to each other:
\[ 1 - \frac{T_2}{1000} = 1 - \frac{360}{T_2} \]

4. **Simplify the Equation:**
We can simplify this equation by eliminating the 1 from both sides:
\[ -\frac{T_2}{1000} = -\frac{360}{T_2} \]
Multiplying both sides by -1 gives:
\[ \frac{T_2}{1000} = \frac{360}{T_2} \]

5. **Cross-Multiply:**
Cross-multiplying gives:
\[ T_2^2 = 360 \times 1000 \]

6. **Calculate \( T_2 \):**
Now, calculate \( T_2 \):
\[ T_2^2 = 360000 \]
Taking the square root of both sides:
\[ T_2 = \sqrt{360000} = 600 \, K \]

### Final Answer:
The temperature \( T_2 \) is \( 600 \, K \).

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