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

Explanation

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

### Step 1: Define the heat capacities
Let \( MCA \), \( MCB \), and \( MCC \) be the heat capacities of liquids A, B, and C respectively. Since the masses are equal, we can express the heat capacities as:
- \( MCA = m \cdot c_A \)
- \( MCB = m \cdot c_B \)
- \( MCC = m \cdot c_C \)

### Step 2: Set up the equations for mixing A and B
When liquids A and B are mixed, the heat lost by B equals the heat gained by A. This can be expressed as:
\[ MCB \cdot (T_B - T_{AB}) = MCA \cdot (T_{AB} - T_A) \]
Substituting the known values:
- \( T_A = 12^\circ C \)
- \( T_B = 19^\circ C \)
- \( T_{AB} = 16^\circ C \)

This gives:
\[ MCB \cdot (19 - 16) = MCA \cdot (16 - 12) \]
\[ MCB \cdot 3 = MCA \cdot 4 \]
Rearranging gives:
\[ \frac{MCA}{MCB} = \frac{3}{4} \quad \text{(Equation 1)} \]

### Step 3: Set up the equations for mixing B and C
Similarly, when liquids B and C are mixed, the heat lost by C equals the heat gained by B:
\[ MCC \cdot (T_C - T_{BC}) = MCB \cdot (T_{BC} - T_B) \]
Substituting the known values:
- \( T_C = 28^\circ C \)
- \( T_{BC} = 23^\circ C \)

This gives:
\[ MCC \cdot (28 - 23) = MCB \cdot (23 - 19) \]
\[ MCC \cdot 5 = MCB \cdot 4 \]
Rearranging gives:
\[ \frac{MCC}{MCB} = \frac{4}{5} \quad \text{(Equation 2)} \]

### Step 4: Express \( MCA \) and \( MCC \) in terms of \( MCB \)
From Equation 1:
\[ MCA = \frac{3}{4} MCB \]
From Equation 2:
\[ MCC = \frac{4}{5} MCB \]

### Step 5: Set up the equation for mixing A and C
When liquids A and C are mixed, the heat lost by C equals the heat gained by A:
\[ MCC \cdot (T_C - T_{AC}) = MCA \cdot (T_{AC} - T_A) \]
Substituting the values:
\[ \frac{4}{5} MCB \cdot (28 - T_{AC}) = \frac{3}{4} MCB \cdot (T_{AC} - 12) \]
Since \( MCB \) can be divided from both sides, we simplify to:
\[ \frac{4}{5} (28 - T_{AC}) = \frac{3}{4} (T_{AC} - 12) \]

### Step 6: Solve for \( T_{AC} \)
Cross-multiplying gives:
\[ 4 \cdot 4 (28 - T_{AC}) = 3 \cdot 5 (T_{AC} - 12) \]
\[ 16(28 - T_{AC}) = 15(T_{AC} - 12) \]
Expanding both sides:
\[ 448 - 16T_{AC} = 15T_{AC} - 180 \]
Rearranging gives:
\[ 448 + 180 = 15T_{AC} + 16T_{AC} \]
\[ 628 = 31T_{AC} \]
Thus,
\[ T_{AC} = \frac{628}{31} \approx 20.3^\circ C \]

### Final Answer
The equilibrium temperature when liquids A and C are mixed is approximately \( 20.3^\circ C \).

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