540 g of ice at `0^(@)C` is mixed with 540 g of water at `80^(@)C`. The final temperature of the mixture is

Options

A. `0^(@)C`
B. `40^(@)C`
C. `80^(@)C`
D. Less than `0^(@)C`

Answer

To solve the problem of mixing 540 g of ice at \(0^\circ C\) with 540 g of water at \(80^\circ C\), we need to consider the heat exchange between the ice and the water. The ice will absorb heat to melt and then warm up, while the water will lose heat as it cools down.

### Step-by-Step Solution:

1. **Identify the Heat Required to Melt the Ice:**
The heat required to melt the ice can be calculated using the formula:
\[ Q_{\text{ice}} = m \cdot L_f \]
where:
- \(m = 540 \, \text{g}\) (mass of ice)
- \(L_f = 80 \, \text{cal/g}\) (latent heat of fusion for ice)

Substituting the values:
\[ Q_{\text{ice}} = 540 \, \text{g} \cdot 80 \, \text{cal/g} = 43200 \, \text{cal} \]

2. **Identify the Heat Lost by the Water:**
The heat lost by the water as it cools down from \(80^\circ C\) to \(0^\circ C\) can be calculated using:
\[ Q_{\text{water}} = m \cdot c \cdot \Delta T \]
where:
- \(m = 540 \, \text{g}\) (mass of water)
- \(c = 1 \, \text{cal/g}^\circ C\) (specific heat of water)
- \(\Delta T = 80^\circ C - 0^\circ C = 80^\circ C\)

Substituting the values:
\[ Q_{\text{water}} = 540 \, \text{g} \cdot 1 \, \text{cal/g}^\circ C \cdot 80^\circ C = 43200 \, \text{cal} \]

3. **Compare Heat Gained and Heat Lost:**
- Heat required to melt the ice: \(43200 \, \text{cal}\)
- Heat lost by the water: \(43200 \, \text{cal}\)

Since the heat lost by the water is equal to the heat gained by the ice, all the ice will melt, and the final temperature of the mixture will be \(0^\circ C\).

4. **Conclusion:**
The final temperature of the mixture is:
\[ \text{Final Temperature} = 0^\circ C \]