`0.1 m^(3)` of water at `80^(@)C` is mixed with `0.3m^(3)` of water at `60^(@)C`. The final temperature 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 when `0.1 m³` of water at `80°C` is mixed with `0.3 m³` of water at `60°C`, we can use the principle of conservation of energy, which states that the heat lost by the hot water will equal the heat gained by the cold water.

### Step-by-Step Solution:

1. **Identify the volumes and temperatures**:
- Volume of hot water (V1) = `0.1 m³`, Temperature (T1) = `80°C`
- Volume of cold water (V2) = `0.3 m³`, Temperature (T2) = `60°C`

2. **Convert volumes to masses**:
Since the density of water is approximately `1000 kg/m³`, we can calculate the masses:
- Mass of hot water (m1) = V1 × density = `0.1 m³ × 1000 kg/m³ = 100 kg`
- Mass of cold water (m2) = V2 × density = `0.3 m³ × 1000 kg/m³ = 300 kg`

3. **Set up the heat transfer equation**:
According to the principle of calorimetry:
\[ \text{Heat lost by hot water} = \text{Heat gained by cold water} \]
This can be expressed as:
\[ m_1 \cdot c \cdot (T_1 - T_f) = m_2 \cdot c \cdot (T_f - T_2) \]
where \( c \) is the specific heat capacity of water (which cancels out since it's the same for both).

4. **Substituting values**:
\[ 100 \cdot (80 - T_f) = 300 \cdot (T_f - 60) \]

5. **Expand and simplify the equation**:
\[ 8000 - 100T_f = 300T_f - 18000 \]
Rearranging gives:
\[ 8000 + 18000 = 300T_f + 100T_f \]
\[ 26000 = 400T_f \]

6. **Solve for \( T_f \)**:
\[ T_f = \frac{26000}{400} = 65°C \]

### Final Answer:
The final temperature of the mixture is **65°C**.