The temperature of equal masses of three different liquids A,B and C are `12^@C, 19^@C and 28^@C` respectively. The temperature when A and B are mixed is `16^@C` and when B and C are mixed it is `23^@C`. What should be the temperature when A and C are mixed?

Options

A. `18.2 .^(@)C`
B. `22 .^(@)C`
C. `20.3 .^(@)C`
D. `24.2 .^(@)C`

Explanation

To solve the problem, we will use the concept of heat exchange and the principle of conservation of energy. When two liquids are mixed, the heat lost by the hotter liquid is equal to the heat gained by the cooler liquid.

### Step-by-Step Solution:

1. **Identify the Temperatures**:
- Let the temperatures of the liquids be:
- Liquid A: \( T_A = 12^\circ C \)
- Liquid B: \( T_B = 19^\circ C \)
- Liquid C: \( T_C = 28^\circ C \)

2. **Mixing A and B**:
- When liquids A and B are mixed, the final temperature \( T_{AB} = 16^\circ C \).
- Using the formula for heat exchange:
\[ m C_A (T_{AB} - T_A) = m C_B (T_B - T_{AB}) \]
- Since the masses are equal, we can cancel \( m \):
\[ C_A (16 - 12) = C_B (19 - 16) \]
- Simplifying gives:
\[ 4 C_A = 3 C_B \quad \text{(Equation 1)} \]

3. **Mixing B and C**:
- When liquids B and C are mixed, the final temperature \( T_{BC} = 23^\circ C \).
- Again using the heat exchange formula:
\[ m C_B (T_{BC} - T_B) = m C_C (T_C - T_{BC}) \]
- Canceling \( m \):
\[ C_B (23 - 19) = C_C (28 - 23) \]
- Simplifying gives:
\[ 4 C_B = 5 C_C \quad \text{(Equation 2)} \]

4. **Express \( C_A \) in terms of \( C_C \)**:
- From Equation 1, we can express \( C_A \) in terms of \( C_B \):
\[ C_A = \frac{3}{4} C_B \]
- From Equation 2, we can express \( C_B \) in terms of \( C_C \):
\[ C_B = \frac{5}{4} C_C \]
- Substituting \( C_B \) into the equation for \( C_A \):
\[ C_A = \frac{3}{4} \left(\frac{5}{4} C_C\right) = \frac{15}{16} C_C \quad \text{(Equation 3)} \]

5. **Mixing A and C**:
- Now, we need to find the temperature \( T \) when A and C are mixed:
\[ m C_A (T - T_A) = m C_C (T_C - T) \]
- Canceling \( m \):
\[ C_A (T - 12) = C_C (28 - T) \]
- Substitute \( C_A \) from Equation 3:
\[ \frac{15}{16} C_C (T - 12) = C_C (28 - T) \]
- Cancel \( C_C \) (assuming \( C_C \neq 0 \)):
\[ \frac{15}{16} (T - 12) = 28 - T \]

6. **Solving for \( T \)**:
- Multiply through by 16 to eliminate the fraction:
\[ 15(T - 12) = 16(28 - T) \]
- Expanding both sides:
\[ 15T - 180 = 448 - 16T \]
- Combine like terms:
\[ 15T + 16T = 448 + 180 \]
\[ 31T = 628 \]
- Divide by 31:
\[ T = \frac{628}{31} \approx 20.26^\circ C \]

### Final Answer:
The temperature when liquids A and C are mixed is approximately \( 20.26^\circ C \).