The temperature of three different liquids `A,B` and `C` are `14^(@)C, 24^(@)C` and `34^(@)C` respectively. When equal masses of `A` and `B` are mixed, the temperature of the mixture is `20^(@)C`. When equal masses of `B` and `C` are mixed, the temperature of the mixture is `31^(@)C`. Supposing equal masses of `A` and `C` were mixed, what would be the temperature of the mixture?

Answer

To solve the problem, we will use the principle of calorimetry, which states that the heat gained by one substance is equal to the heat lost by another when they are mixed.

### Step-by-Step Solution:

1. **Identify the Temperatures**:
- Let the temperatures of liquids A, B, and C be \( T_A = 14^\circ C \), \( T_B = 24^\circ C \), and \( T_C = 34^\circ C \).

2. **Mixing A and B**:
- When equal masses of A and B are mixed, the final temperature \( T_{AB} = 20^\circ C \).
- Using the heat transfer equation:
\[ \text{Heat lost by B} = \text{Heat gained by A} \]
\[ m c_B (T_B - T_{AB}) = m c_A (T_{AB} - T_A) \]
- Canceling \( m \) (mass) from both sides, we have:
\[ c_B (T_B - T_{AB}) = c_A (T_{AB} - T_A) \]
- Substituting the values:
\[ c_B (24 - 20) = c_A (20 - 14) \]
\[ 4c_B = 6c_A \quad \Rightarrow \quad c_A = \frac{2}{3} c_B \quad \text{(1)} \]

3. **Mixing B and C**:
- When equal masses of B and C are mixed, the final temperature \( T_{BC} = 31^\circ C \).
- Again, using the heat transfer equation:
\[ m c_C (T_C - T_{BC}) = m c_B (T_{BC} - T_B) \]
- Canceling \( m \) gives:
\[ c_C (T_C - T_{BC}) = c_B (T_{BC} - T_B) \]
- Substituting the values:
\[ c_C (34 - 31) = c_B (31 - 24) \]
\[ 3c_C = 7c_B \quad \Rightarrow \quad c_C = \frac{7}{3} c_B \quad \text{(2)} \]

4. **Mixing A and C**:
- Now, we need to find the final temperature \( T_{AC} \) when equal masses of A and C are mixed.
- Using the heat transfer equation:
\[ m c_C (T_C - T_{AC}) = m c_A (T_{AC} - T_A) \]
- Canceling \( m \) gives:
\[ c_C (T_C - T_{AC}) = c_A (T_{AC} - T_A) \]
- Substituting the values:
\[ c_C (34 - T_{AC}) = c_A (T_{AC} - 14) \]
- Using equations (1) and (2):
\[ \frac{7}{3} c_B (34 - T_{AC}) = \frac{2}{3} c_B (T_{AC} - 14) \]
- Canceling \( c_B \) and multiplying through by 3:
\[ 7(34 - T_{AC}) = 2(T_{AC} - 14) \]
- Expanding both sides:
\[ 238 - 7T_{AC} = 2T_{AC} - 28 \]
- Rearranging gives:
\[ 238 + 28 = 9T_{AC} \]
\[ 266 = 9T_{AC} \]
\[ T_{AC} = \frac{266}{9} \approx 29.56^\circ C \]

5. **Final Result**:
- The final temperature when equal masses of A and C are mixed is approximately \( 29.56^\circ C \).