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. `20.2^@C`
D. `24.2^@C`

Answer

To find the temperature when liquids A and C are mixed, we will use the principle of heat exchange and the information provided about the mixtures of A & B and B & C.

### Step-by-Step Solution:

1. **Identify the Given Temperatures:**
- Temperature of liquid A, \( T_A = 12^\circ C \)
- Temperature of liquid B, \( T_B = 19^\circ C \)
- Temperature of liquid C, \( T_C = 28^\circ C \)
- Temperature when A and B are mixed, \( T_{AB} = 16^\circ C \)
- Temperature when B and C are mixed, \( T_{BC} = 23^\circ C \)

2. **Set Up the Heat Exchange Equations:**
- For the mixture of A and B:
\[ m C_A (T_{AB} - T_A) = m C_B (T_B - T_{AB}) \]
where \( m \) is the mass of the liquids and \( C_A, C_B \) are their specific heats.
- This simplifies to:
\[ C_A (16 - 12) = C_B (19 - 16) \]
\[ 4 C_A = 3 C_B \quad \text{(Equation 1)} \]

3. **For the mixture of B and C:**
- Set up the equation:
\[ m C_B (T_{BC} - T_B) = m C_C (T_C - T_{BC}) \]
- This simplifies to:
\[ C_B (23 - 19) = C_C (28 - 23) \]
\[ 4 C_B = 5 C_C \quad \text{(Equation 2)} \]

4. **Express \( C_A \) and \( C_B \) in terms of \( C_C \):**
- From Equation 1:
\[ C_A = \frac{3}{4} C_B \]
- From Equation 2:
\[ C_B = \frac{5}{4} C_C \]
- Substitute \( C_B \) into the equation for \( C_A \):
\[ C_A = \frac{3}{4} \left(\frac{5}{4} C_C\right) = \frac{15}{16} C_C \]

5. **Set Up the Equation for A and C:**
- Now, we can find the temperature when A and C are mixed:
\[ m C_A (T - T_A) = m C_C (T_C - T) \]
- This simplifies to:
\[ C_A (T - 12) = C_C (28 - T) \]
- Substitute \( C_A = \frac{15}{16} C_C \):
\[ \frac{15}{16} C_C (T - 12) = C_C (28 - T) \]
- Cancel \( C_C \) (assuming \( C_C \neq 0 \)):
\[ \frac{15}{16} (T - 12) = 28 - T \]

6. **Solve for \( T \):**
- Multiply through by 16 to eliminate the fraction:
\[ 15(T - 12) = 16(28 - T) \]
\[ 15T - 180 = 448 - 16T \]
\[ 15T + 16T = 448 + 180 \]
\[ 31T = 628 \]
\[ T = \frac{628}{31} = 20.16^\circ C \]

### Final Answer:
The temperature when liquids A and C are mixed is approximately \( 20.16^\circ C \).