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. `30g`
B. `80g`
C. `1600g`
D. `150g`

Explanation

To solve the problem of how much ice will melt when 80 g of water at 30°C is poured onto a block of ice at 0°C, we can follow these steps:

### Step 1: Calculate the heat lost by the water
The heat lost by the water as it cools from 30°C to 0°C can be calculated using the formula:

\[ Q_{\text{lost}} = m \cdot c \cdot \Delta T \]

Where:
- \( m \) = mass of the water = 80 g
- \( c \) = specific heat capacity of water = 1 cal/g°C
- \( \Delta T \) = change in temperature = (30 - 0)°C = 30°C

Substituting the values:

\[ Q_{\text{lost}} = 80 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 30 \, \text{°C} = 2400 \, \text{cal} \]

### Step 2: Calculate the heat required to melt the ice
Let \( m \) be the mass of ice that melts. The heat required to melt the ice can be calculated using the formula:

\[ Q_{\text{gained}} = m \cdot L_f \]

Where:
- \( L_f \) = latent heat of fusion of ice = 80 cal/g

So, we have:

\[ Q_{\text{gained}} = m \cdot 80 \, \text{cal/g} \]

### Step 3: Set the heat lost equal to the heat gained
Since the heat lost by the water is equal to the heat gained by the ice, we can set the two equations equal to each other:

\[ 2400 \, \text{cal} = m \cdot 80 \, \text{cal/g} \]

### Step 4: Solve for \( m \)
Now, we can solve for \( m \):

\[ m = \frac{2400 \, \text{cal}}{80 \, \text{cal/g}} = 30 \, \text{g} \]

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

---