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 g
B. 80 g
C. 1600 g
D. 150 g

Answer

To solve the problem, we need to calculate how much ice melts when 80 grams of water at 30°C is poured onto it. We'll use the principle of conservation of energy, where the heat lost by the water will equal the heat gained by the ice.

### Step-by-Step Solution:

1. **Identify the heat lost by the water**:
The heat lost by the water can be calculated using the formula:
\[ Q = m \cdot s \cdot \Delta T \]
where:
- \( Q \) is the heat lost,
- \( m \) is the mass of the water (80 g),
- \( s \) is the specific heat capacity of water (1 cal/g°C),
- \( \Delta T \) is the change in temperature (30°C - 0°C = 30°C).

Plugging in the values:
\[ Q = 80 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (30°C - 0°C) = 80 \cdot 1 \cdot 30 = 2400 \, \text{calories} \]

2. **Identify the heat gained by the ice**:
The heat gained by the ice as it melts can be calculated using the formula:
\[ Q = m \cdot L \]
where:
- \( m \) is the mass of the ice that melts,
- \( L \) is the latent heat of fusion of ice (approximately 80 cal/g).

Setting the heat lost by the water equal to the heat gained by the ice:
\[ 2400 \, \text{calories} = m \cdot 80 \, \text{cal/g} \]

3. **Solve for the mass of ice that melts**:
Rearranging the equation to solve for \( m \):
\[ m = \frac{2400 \, \text{calories}}{80 \, \text{cal/g}} = 30 \, \text{g} \]

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