Cascaded Carnot engine is an arrangement in which heat sink of one engine is source for other. If high temperature for one engine is `T_1`, low temperature for other engine is `T_2` (Assume work done by both engine is same) Calculate lower temperature of first engine

Options

A. `(2T_1T_2)/(T_1+T_2)`
B. `(T_1+T_2)/2`
C. 0
D. `sqrt(T_1T_2)`

Explanation

To solve the problem of finding the lower temperature of the first Carnot engine in a cascaded arrangement, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
We have two Carnot engines in a cascaded arrangement. The first engine (Engine 1) has a high temperature \( T_1 \) and a lower temperature \( T \) (which we need to find). The second engine (Engine 2) has a low temperature \( T_2 \) and takes the heat rejected from Engine 1 as its heat source.

2. **Work Done by the Engines**:
- The work done by Engine 1 is given by:
\[ W_1 = Q_1 - Q \]
where \( Q_1 \) is the heat absorbed from the high-temperature reservoir and \( Q \) is the heat rejected to the lower temperature reservoir (which is also the heat source for Engine 2).
- The work done by Engine 2 is given by:
\[ W_2 = Q - Q_2 \]
where \( Q_2 \) is the heat rejected by Engine 2.

3. **Equating the Work Done**:
Since the work done by both engines is the same, we have:
\[ W_1 = W_2 \]
Thus, we can write:
\[ Q_1 - Q = Q - Q_2 \]

4. **Rearranging the Equation**:
Rearranging the above equation gives:
\[ Q_1 + Q_2 = 2Q \]

5. **Using Carnot's Theorem**:
For a Carnot engine, the efficiency is given by:
\[ \eta = 1 - \frac{T_{low}}{T_{high}} \]
Therefore, for Engine 1:
\[ \frac{Q_1}{Q} = \frac{T_1}{T} \]
And for Engine 2:
\[ \frac{Q}{Q_2} = \frac{T}{T_2} \]

6. **Substituting into the Work Equation**:
From the work equation \( Q_1 + Q_2 = 2Q \), we can express \( Q_1 \) and \( Q_2 \) in terms of \( Q \):
\[ Q_1 = \frac{T_1}{T} Q \quad \text{and} \quad Q_2 = \frac{T}{T_2} Q \]

7. **Substituting into the Equation**:
Substituting these into the equation:
\[ \frac{T_1}{T} Q + \frac{T}{T_2} Q = 2Q \]
Dividing through by \( Q \) (assuming \( Q \neq 0 \)):
\[ \frac{T_1}{T} + \frac{T}{T_2} = 2 \]

8. **Solving for T**:
Multiply through by \( T \) to eliminate the fractions:
\[ T_1 + \frac{T^2}{T_2} = 2T \]
Rearranging gives:
\[ T^2 - 2T + T_1 T_2 = 0 \]

9. **Using the Quadratic Formula**:
We can solve this quadratic equation using the quadratic formula:
\[ T = \frac{2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot T_1 T_2}}{2 \cdot 1} \]
Simplifying gives:
\[ T = 1 \pm \sqrt{1 - T_1 T_2} \]

10. **Final Result**:
The lower temperature \( T \) of the first engine is:
\[ T = \frac{T_1 + T_2}{2} \]