Equal masses of three liquids A, B and C have temperature `10^(@)C, 25^(@)C` and `40^(@)c` respectively. If A and B are mixed, the mixture has a temperature of `15^(@)C`. If B and C are mixed, the mixture has a temperature of `30^(@)C`, if A and C are mixed will have a temperature of

Options

A. `16^(@)C`
B. `20^(@)C`
C. `25^(@)C`
D. `29^(@)C`

Answer

To solve the problem, we need to determine the final temperature when equal masses of liquids A and C are mixed. We will use the principle of conservation of energy, which states that the heat lost by the hotter liquid will be equal to the heat gained by the cooler liquid.

### Step-by-Step Solution:

1. **Identify the Initial Temperatures and Specific Heats:**
- Let the specific heats of liquids A, B, and C be \( s_A, s_B, \) and \( s_C \) respectively.
- The initial temperatures are:
- \( T_A = 10^\circ C \)
- \( T_B = 25^\circ C \)
- \( T_C = 40^\circ C \)

2. **Mixing A and B:**
- When A and B are mixed, the final temperature \( T_{AB} = 15^\circ C \).
- Using the heat balance:
\[ m \cdot s_A \cdot (T_{AB} - T_A) = m \cdot s_B \cdot (T_B - T_{AB}) \]
- Substituting the values:
\[ m \cdot s_A \cdot (15 - 10) = m \cdot s_B \cdot (25 - 15) \]
- This simplifies to:
\[ 5s_A = 10s_B \quad \Rightarrow \quad s_A = 2s_B \quad \text{(Equation 1)} \]

3. **Mixing B and C:**
- When B and C are mixed, the final temperature \( T_{BC} = 30^\circ C \).
- Using the heat balance:
\[ m \cdot s_B \cdot (T_{BC} - T_B) = m \cdot s_C \cdot (T_C - T_{BC}) \]
- Substituting the values:
\[ m \cdot s_B \cdot (30 - 25) = m \cdot s_C \cdot (40 - 30) \]
- This simplifies to:
\[ 5s_B = 10s_C \quad \Rightarrow \quad s_B = 2s_C \quad \text{(Equation 2)} \]

4. **Relating Specific Heats:**
- From Equation 1, we have:
\[ s_A = 2s_B \]
- From Equation 2, substituting \( s_B \):
\[ s_A = 2(2s_C) = 4s_C \]

5. **Mixing A and C:**
- Let the final temperature when A and C are mixed be \( T \).
- Using the heat balance:
\[ m \cdot s_A \cdot (T - T_A) = m \cdot s_C \cdot (T_C - T) \]
- Substituting the values:
\[ m \cdot (4s_C) \cdot (T - 10) = m \cdot s_C \cdot (40 - T) \]
- Canceling \( m \) and \( s_C \) (assuming \( s_C \neq 0 \)):
\[ 4(T - 10) = 40 - T \]
- Expanding and rearranging:
\[ 4T - 40 = 40 - T \]
\[ 5T = 80 \quad \Rightarrow \quad T = 16^\circ C \]

### Final Answer:
The final temperature when liquids A and C are mixed is \( T = 16^\circ C \).