How many calories are required to change one gram of `0^(@)C` ice to `100^(@)C` steam ? The latent heat of fusion is 80 cal `//` g and the latent heat of vaporization is `540 ` cal `//` g. The specific heat of water is `1.00 cal //` (g K ) .

Options

A. 100 cal
B. 540 cal
C. 620 cal
D. 720 cal

Answer

To calculate the total calories required to change 1 gram of ice at 0°C to steam at 100°C, we will go through the process step by step.

### Step 1: Melting Ice to Water
The first step is to convert ice at 0°C to water at 0°C. This requires the latent heat of fusion.

- **Latent heat of fusion** = 80 cal/g
- For 1 gram of ice:
\[ \text{Energy required} = \text{mass} \times \text{latent heat of fusion} = 1 \, \text{g} \times 80 \, \text{cal/g} = 80 \, \text{cal} \]

### Step 2: Heating Water from 0°C to 100°C
Next, we need to heat the water from 0°C to 100°C. This requires the specific heat of water.

- **Specific heat of water** = 1 cal/(g·K)
- Temperature change = 100°C - 0°C = 100 K
- For 1 gram of water:
\[ \text{Energy required} = \text{mass} \times \text{specific heat} \times \text{temperature change} = 1 \, \text{g} \times 1 \, \text{cal/(g·K)} \times 100 \, \text{K} = 100 \, \text{cal} \]

### Step 3: Vaporizing Water to Steam
Finally, we need to convert water at 100°C to steam at 100°C. This requires the latent heat of vaporization.

- **Latent heat of vaporization** = 540 cal/g
- For 1 gram of water:
\[ \text{Energy required} = \text{mass} \times \text{latent heat of vaporization} = 1 \, \text{g} \times 540 \, \text{cal/g} = 540 \, \text{cal} \]

### Step 4: Total Energy Calculation
Now, we sum up all the energies calculated in the previous steps to find the total energy required.

\[ \text{Total energy} = \text{Energy for melting} + \text{Energy for heating} + \text{Energy for vaporization} \]
\[ \text{Total energy} = 80 \, \text{cal} + 100 \, \text{cal} + 540 \, \text{cal} = 720 \, \text{cal} \]

Thus, the total calories required to change 1 gram of 0°C ice to 100°C steam is **720 calories**.