The fraction of ice that melts by mixing equal masses of ice at `- 10^@C` and water at `60^@C` is

Options

A. `(6)/(11)`
B. `(11)/(16)`
C. `(5)/(16)`
D. `(11)/(15)`

Explanation

To solve the problem of finding the fraction of ice that melts when mixing equal masses of ice at -10°C and water at 60°C, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of ice (m) = Mass of water (m)
- Initial temperature of ice (T_ice) = -10°C
- Initial temperature of water (T_water) = 60°C
- Latent heat of fusion of ice (L_f) = 80 cal/g
- Specific heat of water (c_water) = 1 cal/g°C
- Specific heat of ice (c_ice) = 0.5 cal/g°C

2. **Calculate the Heat Lost by Water:**
The water will lose heat as it cools down to 0°C. The heat lost (Q_water) can be calculated using the formula:
\[ Q_{water} = m \cdot c_{water} \cdot (T_{initial} - T_{final}) = m \cdot 1 \cdot (60 - 0) = 60m \]

3. **Calculate the Heat Gained by Ice:**
The ice will first warm up from -10°C to 0°C and then some of it will melt. The heat gained by the ice (Q_ice) can be expressed as:
\[ Q_{ice} = m \cdot c_{ice} \cdot (0 - (-10)) + m' \cdot L_f \]
Here, \(m'\) is the mass of ice that melts. The first part of the equation is the heat required to warm the ice:
\[ Q_{ice} = m \cdot 0.5 \cdot 10 + m' \cdot 80 = 5m + 80m' \]

4. **Set Up the Heat Balance Equation:**
At thermal equilibrium, the heat lost by the water equals the heat gained by the ice:
\[ Q_{water} = Q_{ice} \]
Substituting the expressions we derived:
\[ 60m = 5m + 80m' \]

5. **Solve for the Mass of Ice that Melts (m'):**
Rearranging the equation gives:
\[ 60m - 5m = 80m' \]
\[ 55m = 80m' \]
\[ m' = \frac{55}{80}m = \frac{11}{16}m \]

6. **Calculate the Fraction of Ice that Melts:**
The fraction of ice that melts is given by:
\[ \text{Fraction melted} = \frac{m'}{m} = \frac{\frac{11}{16}m}{m} = \frac{11}{16} \]

### Final Answer:
The fraction of ice that melts is \(\frac{11}{16}\).

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