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 temperature when liquids A and C are mixed, we can use the information provided about the mixtures of A and B, and B and C. Let's go through the solution step by step.

### Step 1: Understand the given data
- Temperature of liquid A (T_A) = 12°C
- Temperature of liquid B (T_B) = 19°C
- Temperature of liquid C (T_C) = 28°C
- Mixture temperature of A and B (T_AB) = 16°C
- Mixture temperature of B and C (T_BC) = 23°C

### Step 2: Set up the equations for A and B
When A and B are mixed, the heat lost by B is equal to the heat gained by A. The equation can be set up as follows:

\[ m C_A (T_{AB} - T_A) = m C_B (T_B - T_{AB}) \]

Where:
- \(C_A\) is the specific heat of A
- \(C_B\) is the specific heat of B

Substituting the known values:

\[ C_A (16 - 12) = C_B (19 - 16) \]

This simplifies to:

\[ 4C_A = 3C_B \quad \text{(1)} \]

### Step 3: Set up the equations for B and C
Similarly, for the mixture of B and C:

\[ m C_B (T_{BC} - T_B) = m C_C (T_C - T_{BC}) \]

Substituting the known values:

\[ C_B (23 - 19) = C_C (28 - 23) \]

This simplifies to:

\[ 4C_B = 5C_C \quad \text{(2)} \]

### Step 4: Express \(C_B\) in terms of \(C_C\)
From equation (2):

\[ C_B = \frac{5}{4}C_C \]

### Step 5: Substitute \(C_B\) in equation (1)
Now substitute \(C_B\) from equation (2) into equation (1):

\[ 4C_A = 3 \left(\frac{5}{4}C_C\right) \]

This simplifies to:

\[ 4C_A = \frac{15}{4}C_C \]

Multiplying both sides by 4:

\[ 16C_A = 15C_C \quad \text{(3)} \]

### Step 6: Set up the equation for A and C
Now we need to find the temperature when A and C are mixed:

\[ m C_A (T - T_A) = m C_C (T_C - T) \]

Substituting the known values:

\[ C_A (T - 12) = C_C (28 - T) \]

### Step 7: Substitute \(C_A\) from equation (3)
From equation (3), we can express \(C_A\) in terms of \(C_C\):

\[ C_A = \frac{15}{16}C_C \]

Substituting this into the equation for A and C:

\[ \frac{15}{16}C_C (T - 12) = C_C (28 - T) \]

### Step 8: Cancel \(C_C\) and solve for T
Assuming \(C_C \neq 0\), we can cancel \(C_C\):

\[ \frac{15}{16}(T - 12) = 28 - T \]

Multiplying through by 16 to eliminate the fraction:

\[ 15(T - 12) = 16(28 - T) \]

Expanding both sides:

\[ 15T - 180 = 448 - 16T \]

### Step 9: Combine like terms
Adding \(16T\) to both sides:

\[ 31T - 180 = 448 \]

Adding 180 to both sides:

\[ 31T = 628 \]

### Step 10: Solve for T
Dividing both sides by 31:

\[ T = \frac{628}{31} \approx 20.26°C \]

### Final Answer
The temperature when liquids A and C are mixed is approximately **20.26°C**.

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