10 g of ice at `0^(@)C` is mixed with m mass of water at `50^(@)C`. What is minimum value of m so that ice melts completely. (L = 80 cal/g and s = 1 cal/`g-.^(@)C`)

Options

A. 32 g
B. 20 g
C. 40 g
D. 16 g

Answer

To solve the problem of determining the minimum mass \( m \) of water at \( 50^\circ C \) needed to completely melt \( 10 \, g \) of ice at \( 0^\circ C \), we can follow these steps:

### Step 1: Calculate the heat required to melt the ice
The heat required to melt the ice can be calculated using the formula:
\[ Q_{\text{ice}} = m_{\text{ice}} \times L \]
Where:
- \( m_{\text{ice}} = 10 \, g \) (mass of ice)
- \( L = 80 \, \text{cal/g} \) (latent heat of fusion)

Substituting the values:
\[ Q_{\text{ice}} = 10 \, g \times 80 \, \text{cal/g} = 800 \, \text{cal} \]

### Step 2: Calculate the heat provided by the water
The heat provided by the water as it cools down from \( 50^\circ C \) to \( 0^\circ C \) can be calculated using the formula:
\[ Q_{\text{water}} = m \times s \times \Delta T \]
Where:
- \( m \) is the mass of water (unknown)
- \( s = 1 \, \text{cal/g} \, ^\circ C \) (specific heat of water)
- \( \Delta T = 50^\circ C - 0^\circ C = 50^\circ C \)

Substituting the values:
\[ Q_{\text{water}} = m \times 1 \, \text{cal/g} \times 50^\circ C = 50m \, \text{cal} \]

### Step 3: Set the heat gained by ice equal to the heat lost by water
For the ice to melt completely, the heat gained by the ice must equal the heat lost by the water:
\[ Q_{\text{ice}} = Q_{\text{water}} \]
Substituting the expressions we found:
\[ 800 \, \text{cal} = 50m \, \text{cal} \]

### Step 4: Solve for \( m \)
To find \( m \), we rearrange the equation:
\[ m = \frac{800 \, \text{cal}}{50 \, \text{cal}} = 16 \, g \]

### Conclusion
The minimum mass \( m \) of water required to completely melt \( 10 \, g \) of ice at \( 0^\circ C \) is \( 16 \, g \).

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