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?

Explanation

To find the equilibrium temperature when equal masses of liquids A and C are mixed, we can follow these steps:

### Step 1: Define the known temperatures and heat capacities
Let:
- Temperature of liquid A, \( T_A = 12^\circ C \)
- Temperature of liquid B, \( T_B = 19^\circ C \)
- Temperature of liquid C, \( T_C = 28^\circ C \)

Let the heat capacities of the liquids be:
- \( mC_A \) for liquid A
- \( mC_B \) for liquid B
- \( mC_C \) for liquid C

### Step 2: Set up the equation for mixing A and B
When A and B are mixed, the heat lost by B equals the heat gained by A:
\[ mC_B (T_B - T_{A+B}) = mC_A (T_{A+B} - T_A) \]
Given that \( T_{A+B} = 16^\circ C \), we can substitute:
\[ mC_B (19 - 16) = mC_A (16 - 12) \]
This simplifies to:
\[ mC_B \cdot 3 = mC_A \cdot 4 \]
Rearranging gives:
\[ \frac{mC_A}{mC_B} = \frac{3}{4} \quad \text{(Equation 1)} \]

### Step 3: Set up the equation for mixing B and C
When B and C are mixed, the heat lost by C equals the heat gained by B:
\[ mC_C (T_C - T_{B+C}) = mC_B (T_{B+C} - T_B) \]
Given that \( T_{B+C} = 23^\circ C \), we can substitute:
\[ mC_C (28 - 23) = mC_B (23 - 19) \]
This simplifies to:
\[ mC_C \cdot 5 = mC_B \cdot 4 \]
Rearranging gives:
\[ \frac{mC_C}{mC_B} = \frac{4}{5} \quad \text{(Equation 2)} \]

### Step 4: Express the heat capacities in terms of \( mC_B \)
From Equation 1:
\[ mC_A = \frac{3}{4} mC_B \]
From Equation 2:
\[ mC_C = \frac{4}{5} mC_B \]

### Step 5: Set up the equation for mixing A and C
When A and C are mixed, the heat lost by C equals the heat gained by A:
\[ mC_C (T_C - T_{A+C}) = mC_A (T_{A+C} - T_A) \]
Substituting the expressions for \( mC_A \) and \( mC_C \):
\[ \left(\frac{4}{5} mC_B\right) (28 - T_{A+C}) = \left(\frac{3}{4} mC_B\right) (T_{A+C} - 12) \]
We can cancel \( mC_B \) from both sides:
\[ \frac{4}{5} (28 - T_{A+C}) = \frac{3}{4} (T_{A+C} - 12) \]

### Step 6: Solve for \( T_{A+C} \)
Cross-multiplying gives:
\[ 4 \cdot 4 (28 - T_{A+C}) = 3 \cdot 5 (T_{A+C} - 12) \]
This simplifies to:
\[ 16(28 - T_{A+C}) = 15(T_{A+C} - 12) \]
Expanding both sides:
\[ 448 - 16T_{A+C} = 15T_{A+C} - 180 \]
Combining like terms:
\[ 448 + 180 = 15T_{A+C} + 16T_{A+C} \]
\[ 628 = 31T_{A+C} \]
Dividing by 31:
\[ T_{A+C} = 20.16^\circ C \]

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

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