80 g of water at `30^@C` are poured on a large block of ice at `0^@C`. The mass of ice that melts is

Options

A. 30 gm
B. 80 gm
C. 1600 gm
D. 150 gm

Answer

To solve the problem of how much ice melts when 80 g of water at 30°C is poured onto it, we can use the principle of conservation of energy, which states that the heat lost by the water will be equal to the heat gained by the ice.

### Step-by-step Solution:

1. **Identify the Given Data:**
- Mass of water (m_water) = 80 g
- Initial temperature of water (T_initial_water) = 30°C
- Final temperature of water (T_final) = 0°C (since it will reach the temperature of ice)
- Specific heat capacity of water (c_water) = 1 cal/g°C (or 4.18 J/g°C)
- Latent heat of fusion of ice (L_ice) = 80 cal/g

2. **Calculate the Heat Lost by the Water:**
The heat lost by the water as it cools down from 30°C to 0°C can be calculated using the formula:
\[ Q_{\text{lost}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{initial water}} - T_{\text{final}}) \]
Substituting the values:
\[ Q_{\text{lost}} = 80 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (30°C - 0°C) = 80 \cdot 1 \cdot 30 = 2400 \, \text{cal} \]

3. **Set Up the Equation for Heat Gained by Ice:**
The heat gained by the ice as it melts can be expressed as:
\[ Q_{\text{gained}} = m_{\text{ice}} \cdot L_{\text{ice}} \]
where \( m_{\text{ice}} \) is the mass of ice that melts.

4. **Equate Heat Lost and Heat Gained:**
Since the heat lost by the water is equal to the heat gained by the ice, we can write:
\[ Q_{\text{lost}} = Q_{\text{gained}} \]
Therefore,
\[ 2400 \, \text{cal} = m_{\text{ice}} \cdot 80 \, \text{cal/g} \]

5. **Solve for the Mass of Ice that Melts:**
Rearranging the equation to find \( m_{\text{ice}} \):
\[ m_{\text{ice}} = \frac{2400 \, \text{cal}}{80 \, \text{cal/g}} = 30 \, \text{g} \]

6. **Conclusion:**
The mass of ice that melts is 30 g.

### Final Answer:
The mass of ice that melts is **30 grams**.