Calculate the amount of heat required to convert 10 g of ice at `0^@C` to 10 g of water at `100^@C`. (Specific heat capacity of water = 1 cal/g/`""^(@)`C)

Options

A. 200 cal
B. 1800 cal
C. 2200 cal
D. 1600 cal

Explanation

To calculate the amount of heat required to convert 10 g of ice at 0°C to 10 g of water at 100°C, we need to consider two steps:

1. **Melting the Ice at 0°C to Water at 0°C**
2. **Heating the Water from 0°C to 100°C**

### Step 1: Melting the Ice

To convert ice at 0°C to water at 0°C, we use the formula:

\[ Q_1 = m \times L \]

Where:
- \( Q_1 \) = heat required to melt the ice
- \( m \) = mass of ice (10 g)
- \( L \) = latent heat of fusion of ice (approximately 80 cal/g)

Substituting the values:

\[ Q_1 = 10 \, \text{g} \times 80 \, \text{cal/g} = 800 \, \text{cal} \]

### Step 2: Heating the Water

Next, we need to heat the water from 0°C to 100°C. We use the formula:

\[ Q_2 = m \times c \times \Delta T \]

Where:
- \( Q_2 \) = heat required to heat the water
- \( m \) = mass of water (10 g)
- \( c \) = specific heat capacity of water (1 cal/g/°C)
- \( \Delta T \) = change in temperature (100°C - 0°C = 100°C)

Substituting the values:

\[ Q_2 = 10 \, \text{g} \times 1 \, \text{cal/g/°C} \times 100 \, \text{°C} = 1000 \, \text{cal} \]

### Total Heat Required

Now, we add the heat required for both steps to find the total heat:

\[ Q_{total} = Q_1 + Q_2 \]

Substituting the values:

\[ Q_{total} = 800 \, \text{cal} + 1000 \, \text{cal} = 1800 \, \text{cal} \]

Thus, the total amount of heat required to convert 10 g of ice at 0°C to 10 g of water at 100°C is **1800 cal**.