2 kg ice at `0^@C` is mixed with 8 kg of water at `20^@C`. The final temperature is

Options

A. `0^@C`
B. `20^@C`
C. `80^@C`
D. none of these

Explanation

To solve the problem of finding the final temperature when 2 kg of ice at 0°C is mixed with 8 kg of water at 20°C, we will use the principle of calorimetry, which states that the heat lost by the warmer substance (water) will be equal to the heat gained by the colder substance (ice).

### Step-by-Step Solution:

1. **Convert Masses to Grams**:
- Mass of ice, \( m_i = 2 \, \text{kg} = 2000 \, \text{g} \)
- Mass of water, \( m_w = 8 \, \text{kg} = 8000 \, \text{g} \)

2. **Identify Specific Heat and Latent Heat**:
- Specific heat of water, \( c_w = 1 \, \text{cal/g°C} \)
- Latent heat of fusion of ice, \( L_f = 80 \, \text{cal/g} \)

3. **Set Up the Heat Transfer Equation**:
- Heat lost by water = Heat gained by ice
- The equation can be expressed as:
\[ m_w \cdot c_w \cdot (T_w - T_f) = m_i \cdot L_f + m_i \cdot c_w \cdot (T_f - T_i) \]
- Where:
- \( T_w = 20°C \) (initial temperature of water)
- \( T_i = 0°C \) (initial temperature of ice)
- \( T_f \) = final temperature

4. **Substituting Values**:
- Substitute the known values into the equation:
\[ 8000 \cdot 1 \cdot (20 - T_f) = 2000 \cdot 80 + 2000 \cdot 1 \cdot (T_f - 0) \]
- This simplifies to:
\[ 8000(20 - T_f) = 160000 + 2000T_f \]

5. **Expand and Rearrange the Equation**:
- Expanding the left side:
\[ 160000 - 8000T_f = 160000 + 2000T_f \]
- Rearranging gives:
\[ 160000 - 160000 = 8000T_f + 2000T_f \]
\[ 0 = 10000T_f \]

6. **Solve for Final Temperature**:
- Dividing both sides by 10000:
\[ T_f = 0°C \]

### Conclusion:
The final temperature when 2 kg of ice at 0°C is mixed with 8 kg of water at 20°C is **0°C**.