Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures, `T_(1) and T_(2)`. The temperature of the hot reservoir of the first engine is `T_(1)` and the temperature of the cold reservoir of the second engine is `T_(2)`. T is temperature of the sink of first engine which is also the source for the second engine. How is T related to `T_(1) and T_(2)`, if both the engines perform equal amount of work?

Options

A. `T=(2T_(1)T_(2))/(T_(1)+T_(2))`
B. `T=(T_(1)+T_(2))/(2)`
C. `T=sqrt(T_(1)T_(2))`
D. `T=0`

Answer

To solve the problem, we need to analyze the two Carnot engines operating in cascade and establish the relationship between the temperatures involved.

### Step-by-Step Solution:

1. **Understanding the Engines**:
- Let Engine 1 operate between temperatures \( T_1 \) (hot reservoir) and \( T \) (cold reservoir).
- Let Engine 2 operate between temperatures \( T \) (hot reservoir) and \( T_2 \) (cold reservoir).

2. **Work Done by Each Engine**:
- The work done by Engine 1, \( W_1 \), can be expressed as:
\[ W_1 = Q_1 - Q_2 \]
where \( Q_1 \) is the heat absorbed from the hot reservoir \( T_1 \) and \( Q_2 \) is the heat rejected to the cold reservoir \( T \).
- The work done by Engine 2, \( W_2 \), can be expressed as:
\[ W_2 = Q_2 - Q_3 \]
where \( Q_3 \) is the heat rejected to the cold reservoir \( T_2 \).

3. **Equating the Work Done**:
- Since it is given that both engines perform equal amounts of work, we have:
\[ W_1 = W_2 \]
Therefore:
\[ Q_1 - Q_2 = Q_2 - Q_3 \]

4. **Rearranging the Equation**:
- Rearranging gives us:
\[ Q_1 + Q_3 = 2Q_2 \]

5. **Expressing Heat in Terms of Temperatures**:
- For Engine 1 (using Carnot's theorem):
\[ \frac{Q_1}{Q_2} = \frac{T_1}{T} \]
- For Engine 2:
\[ \frac{Q_2}{Q_3} = \frac{T}{T_2} \]
- Taking the reciprocal for Engine 2 gives:
\[ \frac{Q_3}{Q_2} = \frac{T_2}{T} \]

6. **Substituting into the Work Equation**:
- From \( Q_1 + Q_3 = 2Q_2 \):
\[ Q_1 = \frac{T_1}{T} Q_2 \]
\[ Q_3 = \frac{T_2}{T} Q_2 \]
- Substituting these into the equation gives:
\[ \frac{T_1}{T} Q_2 + \frac{T_2}{T} Q_2 = 2Q_2 \]

7. **Simplifying**:
- Dividing through by \( Q_2 \) (assuming \( Q_2 \neq 0 \)):
\[ \frac{T_1}{T} + \frac{T_2}{T} = 2 \]
- This simplifies to:
\[ \frac{T_1 + T_2}{T} = 2 \]
- Rearranging gives:
\[ T = \frac{T_1 + T_2}{2} \]

### Final Answer:
Thus, the temperature \( T \) is related to \( T_1 \) and \( T_2 \) as:
\[ T = \frac{T_1 + T_2}{2} \]