`0.1 m^(3)` of water at `80^(@)C` is mixed with `0.3m^(3)` of water at `60^(@)C`. The finial temparature of the mixture is

Options

A. `70^(@)C`
B. `65^(@)C`
C. `60^(@)C`
D. `75^(@)C`

Explanation

To find the final temperature of the mixture of water at different temperatures, we can use the principle of calorimetry, which states that the heat lost by the hotter water will be equal to the heat gained by the cooler water.

### Step-by-Step Solution:

1. **Identify the Masses of Water**:
- The volume of the first body of water (m1) is \(0.1 \, m^3\) at \(80^\circ C\).
- The volume of the second body of water (m2) is \(0.3 \, m^3\) at \(60^\circ C\).
- The density of water (\(\rho\)) is \(1000 \, kg/m^3\).

Using the formula for mass:
\[ m_1 = V_1 \times \rho = 0.1 \, m^3 \times 1000 \, kg/m^3 = 100 \, kg \]
\[ m_2 = V_2 \times \rho = 0.3 \, m^3 \times 1000 \, kg/m^3 = 300 \, kg \]

2. **Set Up the Heat Transfer Equation**:
- Let the final temperature of the mixture be \(T\).
- The heat lost by the first body of water (hot water) is:
\[ Q_{lost} = m_1 \cdot c \cdot (T_1 - T) = 100 \cdot c \cdot (80 - T) \]
- The heat gained by the second body of water (cold water) is:
\[ Q_{gained} = m_2 \cdot c \cdot (T - T_2) = 300 \cdot c \cdot (T - 60) \]

According to the principle of calorimetry:
\[ Q_{lost} = Q_{gained} \]
Therefore:
\[ 100 \cdot c \cdot (80 - T) = 300 \cdot c \cdot (T - 60) \]

3. **Cancel Out the Specific Heat**:
Since the specific heat \(c\) is the same for both bodies of water, we can cancel it out from both sides:
\[ 100 \cdot (80 - T) = 300 \cdot (T - 60) \]

4. **Expand and Rearrange the Equation**:
Expanding both sides:
\[ 8000 - 100T = 300T - 18000 \]
Rearranging gives:
\[ 8000 + 18000 = 300T + 100T \]
\[ 26000 = 400T \]

5. **Solve for \(T\)**:
Dividing both sides by 400:
\[ T = \frac{26000}{400} = 65^\circ C \]

### Final Answer:
The final temperature of the mixture is \(65^\circ C\).