Find the heat required to convert 1 g of ice at 0°0C to vapour at 100°C.
(Latent heat of ice = 80 cal/g and latent heat of vapour = 536 cal/g)

Options

A. 80 cal
B. 536 cal
C. 716 cal
D. None of these

Explanation

To find the heat required to convert 1 g of ice at 0°C to vapor at 100°C, we will break the process down into three stages:

### Step 1: Convert Ice at 0°C to Water at 0°C
- **Formula**: \( Q_1 = m \times L_f \)
- **Where**:
- \( m = 1 \, \text{g} \) (mass of ice)
- \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion for ice)

Calculating:
\[ Q_1 = 1 \, \text{g} \times 80 \, \text{cal/g} = 80 \, \text{cal} \]

### Step 2: Heat Water from 0°C to 100°C
- **Formula**: \( Q_2 = m \times C \times \Delta T \)
- **Where**:
- \( C = 1 \, \text{cal/g°C} \) (specific heat of water)
- \( \Delta T = 100°C - 0°C = 100°C \)

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

### Step 3: Convert Water at 100°C to Vapor at 100°C
- **Formula**: \( Q_3 = m \times L_v \)
- **Where**:
- \( L_v = 536 \, \text{cal/g} \) (latent heat of vaporization)

Calculating:
\[ Q_3 = 1 \, \text{g} \times 536 \, \text{cal/g} = 536 \, \text{cal} \]

### Total Heat Required
Now, we sum up the heat required for all three stages:
\[ Q = Q_1 + Q_2 + Q_3 \]
Calculating:
\[ Q = 80 \, \text{cal} + 100 \, \text{cal} + 536 \, \text{cal} = 716 \, \text{cal} \]

Thus, the total heat required to convert 1 g of ice at 0°C to vapor at 100°C is **716 calories**.

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