`1 gm` of ice at `0^@C` is converted to steam at `100^@C` the amount of heat required will be `(L_("steam") = 536 cal//g)`.

Options

A. 756 cal
B. 12000 cal
C. 716 cal
D. 450 cal

Explanation

To find the total amount of heat required to convert 1 gram of ice at 0°C to steam at 100°C, we need to consider the different stages of the phase change and temperature change involved in this process.

### Step-by-Step Solution:

1. **Convert Ice to Water**:
- The first step is to convert 1 gram of ice at 0°C to water at 0°C. This requires the latent heat of fusion.
- The latent heat of fusion for ice is approximately 80 calories/gram.
- Heat required (Q1) = mass × latent heat of fusion = 1 g × 80 cal/g = **80 calories**.

2. **Heat 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 capacity of water.
- The specific heat capacity of water is approximately 1 cal/g°C.
- The temperature change (ΔT) = 100°C - 0°C = 100°C.
- Heat required (Q2) = mass × specific heat capacity × ΔT = 1 g × 1 cal/g°C × 100°C = **100 calories**.

3. **Convert Water to Steam**:
- Finally, we convert the water at 100°C to steam at 100°C. This requires the latent heat of vaporization.
- The latent heat of vaporization for water is approximately 536 calories/gram.
- Heat required (Q3) = mass × latent heat of vaporization = 1 g × 536 cal/g = **536 calories**.

4. **Total Heat Required**:
- Now, we sum up all the heat required for each stage:
\[ \text{Total Heat (Q)} = Q1 + Q2 + Q3 = 80 \text{ cal} + 100 \text{ cal} + 536 \text{ cal} = 716 \text{ calories}. \]

### Final Answer:
The total amount of heat required to convert 1 gram of ice at 0°C to steam at 100°C is **716 calories**.