When 50 g of liquid P at temperature 80°C is mixed with 100 g of liquid Q at 40°C, the temperature of mixture becomes 60°C. If 25 g of liquid P at 100°C is added to 100 g of liquid Q at 25°C, then the temperature of mixture will be :

Options

A. `70^(@)C`
B. `60^(@)C`
C. `40^(@)C`
D. `50^(@)C`

Answer

To solve the problem, 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 colder liquid.

### Step-by-Step Solution:

1. **Identify the given data:**
- Mass of liquid P (mP) = 25 g
- Initial temperature of liquid P (TP_initial) = 100°C
- Mass of liquid Q (mQ) = 100 g
- Initial temperature of liquid Q (TQ_initial) = 25°C
- Final temperature of the mixture (T_final) = T (unknown)

2. **Define the specific heat capacities:**
- Let the specific heat capacity of liquid P be Cp.
- Let the specific heat capacity of liquid Q be Cq.
- From the previous part of the problem, we know that Cp = 2 * Cq.

3. **Set up the heat transfer equations:**
- Heat lost by liquid P:
\[ Q_{lost} = m_P \cdot C_p \cdot (T_P_{initial} - T_{final}) = 25 \cdot C_p \cdot (100 - T) \]
- Heat gained by liquid Q:
\[ Q_{gained} = m_Q \cdot C_q \cdot (T_{final} - T_Q_{initial}) = 100 \cdot C_q \cdot (T - 25) \]

4. **Equate the heat lost and gained:**
\[ 25 \cdot C_p \cdot (100 - T) = 100 \cdot C_q \cdot (T - 25) \]

5. **Substitute Cp in terms of Cq:**
- Since \( C_p = 2 \cdot C_q \):
\[ 25 \cdot (2 \cdot C_q) \cdot (100 - T) = 100 \cdot C_q \cdot (T - 25) \]

6. **Simplify the equation:**
- Cancel \( C_q \) from both sides (assuming \( C_q \neq 0 \)):
\[ 50 \cdot (100 - T) = 100 \cdot (T - 25) \]

7. **Distribute and rearrange:**
\[ 5000 - 50T = 100T - 2500 \]
\[ 5000 + 2500 = 100T + 50T \]
\[ 7500 = 150T \]

8. **Solve for T:**
\[ T = \frac{7500}{150} = 50°C \]

### Final Answer:
The final temperature of the mixture is **50°C**.

---