Two liquids `A` and `B` are at temperatures of `75^@ C` and `150^@ C` respectively. Their masses are in the ratio of `2 : 3` and specific heats are in the ratio `3 : 4`. The resultant temperature of the mixture, when the above liquids, are mixed (Neglect the water equivalent of container) is

Options

A. `125^@C`
B. `100^@ C`
C. `50^@ C`
D. `150^@ C`

Explanation

To find the resultant temperature when two liquids A and B are mixed, we can use the principle of calorimetry, 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 Given Data:**
- Temperature of liquid A, \( T_A = 75^\circ C \)
- Temperature of liquid B, \( T_B = 150^\circ C \)
- Mass ratio of A to B, \( \frac{M_A}{M_B} = \frac{2}{3} \)
- Specific heat ratio of A to B, \( \frac{S_A}{S_B} = \frac{3}{4} \)

2. **Let the Masses and Specific Heats:**
- Let the mass of liquid A be \( M_A = 2x \) and mass of liquid B be \( M_B = 3x \).
- Let the specific heat of liquid A be \( S_A = 3y \) and specific heat of liquid B be \( S_B = 4y \).

3. **Set Up the Heat Transfer Equation:**
- Heat gained by liquid A:
\[ Q_A = M_A \cdot S_A \cdot (T - T_A) = 2x \cdot 3y \cdot (T - 75) \]
- Heat lost by liquid B:
\[ Q_B = M_B \cdot S_B \cdot (T_B - T) = 3x \cdot 4y \cdot (150 - T) \]

4. **Apply the Principle of Calorimetry:**
- According to the principle, the heat gained by A equals the heat lost by B:
\[ 2x \cdot 3y \cdot (T - 75) = 3x \cdot 4y \cdot (150 - T) \]

5. **Cancel Common Factors:**
- We can cancel \( x \) and \( y \) from both sides:
\[ 6(T - 75) = 12(150 - T) \]

6. **Expand and Rearrange the Equation:**
- Expanding both sides:
\[ 6T - 450 = 1800 - 12T \]
- Rearranging gives:
\[ 6T + 12T = 1800 + 450 \]
\[ 18T = 2250 \]

7. **Solve for T:**
- Dividing both sides by 18:
\[ T = \frac{2250}{18} = 125^\circ C \]

### Final Result:
The resultant temperature of the mixture is \( T = 125^\circ C \).

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