`M` gram of ice at `0^@ C` is mixed with `3 M` gram of water at `80^@ C` then the final temperature is.

Options

A. `30^@ C`
B. `40^@C`
C. `50^@ C`
D. `60^@ C`

Explanation

To solve the problem of finding the final temperature when `M` grams of ice at `0°C` is mixed with `3M` grams of water at `80°C`, we can use the principle of calorimetry, which states that the heat lost by the hot water will be equal to the heat gained by the ice.

### Step-by-Step Solution:

1. **Identify the Heat Gained by Ice**:
- The ice at `0°C` will first absorb heat to melt into water at `0°C`. The latent heat of fusion for ice is `80 cal/g`.
- The heat absorbed by `M` grams of ice to melt is:
\[ Q_{\text{melt}} = M \times 80 \]
- After melting, the resulting water at `0°C` will absorb heat to reach the final temperature `T`. The specific heat of water is `1 cal/g°C`.
- The heat absorbed by this water to reach temperature `T` is:
\[ Q_{\text{heat}} = M \times 1 \times T = M \times T \]
- Therefore, the total heat gained by the ice is:
\[ Q_{\text{ice}} = Q_{\text{melt}} + Q_{\text{heat}} = M \times 80 + M \times T \]

2. **Identify the Heat Lost by Water**:
- The `3M` grams of water at `80°C` will lose heat as it cools down to the final temperature `T`.
- The heat lost by the water is:
\[ Q_{\text{water}} = 3M \times 1 \times (80 - T) = 3M \times (80 - T) \]

3. **Set Up the Equation**:
- According to the principle of calorimetry:
\[ Q_{\text{ice}} = Q_{\text{water}} \]
- Substituting the expressions we found:
\[ M \times 80 + M \times T = 3M \times (80 - T) \]

4. **Simplify the Equation**:
- Divide through by `M` (assuming `M ≠ 0`):
\[ 80 + T = 3(80 - T) \]
- Expanding the right side:
\[ 80 + T = 240 - 3T \]

5. **Combine Like Terms**:
- Rearranging gives:
\[ T + 3T = 240 - 80 \]
\[ 4T = 160 \]

6. **Solve for T**:
- Divide both sides by `4`:
\[ T = 40°C \]

### Final Answer:
The final temperature when `M` grams of ice at `0°C` is mixed with `3M` grams of water at `80°C` is **40°C**.