Equal masses of three liquids A, B and C have temperature `10^(@)C, 25^(@)C` and `40^(@)c` respectively. If A and B are mixed, the mixture has a temperature of `15^(@)C`. If B and C are mixed, the mixture has a temperature of `30^(@)C`, if A and C are mixed will have a temperature of

Options

A. `16^(@)C`
B. `20^(@)C`
C. `25^(@)C`
D. `29^(@)C`

Explanation

To solve the problem, we need to find the equilibrium temperature when equal masses of liquids A and C are mixed. We will use the principle of conservation of energy, which states that the heat lost by the hotter liquid will be equal to the heat gained by the cooler liquid.

### Step-by-Step Solution:

1. **Identify the Given Temperatures**:
- Temperature of liquid A, \( T_A = 10^\circ C \)
- Temperature of liquid B, \( T_B = 25^\circ C \)
- Temperature of liquid C, \( T_C = 40^\circ C \)

2. **Mixing A and B**:
- When A and B are mixed, the final temperature \( T_{AB} = 15^\circ C \).
- Using the heat balance:
\[ m \cdot c_A \cdot (T_{AB} - T_A) = m \cdot c_B \cdot (T_B - T_{AB}) \]
- Substituting the values:
\[ m \cdot c_A \cdot (15 - 10) = m \cdot c_B \cdot (25 - 15) \]
- This simplifies to:
\[ 5c_A = 10c_B \implies c_A = 2c_B \quad \text{(1)} \]

3. **Mixing B and C**:
- When B and C are mixed, the final temperature \( T_{BC} = 30^\circ C \).
- Using the heat balance:
\[ m \cdot c_B \cdot (T_{BC} - T_B) = m \cdot c_C \cdot (T_C - T_{BC}) \]
- Substituting the values:
\[ m \cdot c_B \cdot (30 - 25) = m \cdot c_C \cdot (40 - 30) \]
- This simplifies to:
\[ 5c_B = 10c_C \implies c_B = 2c_C \quad \text{(2)} \]

4. **Relate the Specific Heats**:
- From equation (1): \( c_A = 2c_B \)
- From equation (2): \( c_B = 2c_C \)
- Substitute (2) into (1):
\[ c_A = 2(2c_C) = 4c_C \]

5. **Mixing A and C**:
- Now we need to find the final temperature \( T \) when A and C are mixed.
- Using the heat balance:
\[ m \cdot c_A \cdot (T - T_A) = m \cdot c_C \cdot (T_C - T) \]
- Substituting the values:
\[ m \cdot (4c_C) \cdot (T - 10) = m \cdot c_C \cdot (40 - T) \]
- Cancel \( m \) and \( c_C \):
\[ 4(T - 10) = 40 - T \]
- Expanding and rearranging:
\[ 4T - 40 = 40 - T \]
\[ 4T + T = 80 \]
\[ 5T = 80 \implies T = 16^\circ C \]

### Conclusion:
The equilibrium temperature when liquids A and C are mixed is \( T = 16^\circ C \).