Three liquids A, B and C having same sepcific heats have masses m,2m and 3m. Their temperaures are 0,20 and `3theta` respectively. ltBrgt Q. What is the temperature of mixture, when Aand C all are mixed?

Options

A. `(5)/(2)theta`
B. `(5)/(3)theta`
C. `(7)/(3)theta`
D. `(13)/(5)theta`

Answer

To find the final temperature of the mixture when liquids A and C are mixed, we can use the principle of conservation of energy. The heat lost by liquid A will be equal to the heat gained by liquid C.

### Step-by-Step Solution:

1. **Identify the parameters:**
- Mass of liquid A, \( m_A = m \)
- Mass of liquid C, \( m_C = 3m \)
- Initial temperature of liquid A, \( T_A = 0 \, \text{°C} \)
- Initial temperature of liquid C, \( T_C = 3\theta \, \text{°C} \)
- Specific heat of all liquids, \( s \) (same for A, B, and C)

2. **Set up the heat loss and gain equations:**
- Heat lost by liquid A:
\[ Q_A = m \cdot s \cdot (T_A - T) = m \cdot s \cdot (0 - T) = -m \cdot s \cdot T \]
- Heat gained by liquid C:
\[ Q_C = 3m \cdot s \cdot (T - T_C) = 3m \cdot s \cdot (T - 3\theta) \]

3. **Apply the conservation of energy:**
According to the principle of conservation of energy, the heat lost by A equals the heat gained by C:
\[ Q_C = -Q_A \]
Substituting the expressions for \( Q_A \) and \( Q_C \):
\[ 3m \cdot s \cdot (T - 3\theta) = m \cdot s \cdot T \]

4. **Simplify the equation:**
Divide both sides by \( ms \) (since \( s \) is the same and \( m \) is non-zero):
\[ 3(T - 3\theta) = T \]

5. **Distribute and rearrange:**
\[ 3T - 9\theta = T \]
\[ 3T - T = 9\theta \]
\[ 2T = 9\theta \]

6. **Solve for \( T \):**
\[ T = \frac{9\theta}{2} = 4.5\theta \]

Thus, the final temperature of the mixture when liquids A and C are mixed is:
\[ T = \frac{9\theta}{2} \]