`1` gram of ice at `-10^@ C` is converted to steam at `100^@ C` the amount of heat required is `(S_(ice) = 0.5 cal//g -^(@) C)`
`(L_(v) = 536 cal//g & L_(f) = 80 cal//g,)`.

Options

A. 861 cal
B. 12005 cal
C. 721 cal
D. 455 cal

Explanation

To find the total amount of heat required to convert 1 gram of ice at -10°C to steam at 100°C, we will break down the process into several steps, calculating the heat required for each step.

### Step 1: Heating the Ice from -10°C to 0°C
The first step is to heat the ice from -10°C to 0°C. We can use the formula:
\[ Q_1 = m \cdot S_{ice} \cdot \Delta T \]
Where:
- \( m = 1 \, \text{g} \) (mass of ice)
- \( S_{ice} = 0.5 \, \text{cal/g°C} \) (specific heat of ice)
- \( \Delta T = 0 - (-10) = 10 \, \text{°C} \) (change in temperature)

Calculating \( Q_1 \):
\[ Q_1 = 1 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot 10 \, \text{°C} = 5 \, \text{cal} \]

### Step 2: Melting the Ice at 0°C
Next, we need to melt the ice at 0°C to convert it to water at the same temperature. The heat required for this is given by:
\[ Q_2 = m \cdot L_f \]
Where:
- \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion)

Calculating \( Q_2 \):
\[ Q_2 = 1 \, \text{g} \cdot 80 \, \text{cal/g} = 80 \, \text{cal} \]

### Step 3: Heating the Water from 0°C to 100°C
Now, we heat the water from 0°C to 100°C. The heat required is:
\[ Q_3 = m \cdot S_{water} \cdot \Delta T \]
Where:
- \( S_{water} = 1 \, \text{cal/g°C} \) (specific heat of water)
- \( \Delta T = 100 - 0 = 100 \, \text{°C} \) (change in temperature)

Calculating \( Q_3 \):
\[ Q_3 = 1 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 100 \, \text{°C} = 100 \, \text{cal} \]

### Step 4: Converting Water to Steam at 100°C
Finally, we convert the water at 100°C to steam at the same temperature. The heat required for this is:
\[ Q_4 = m \cdot L_v \]
Where:
- \( L_v = 536 \, \text{cal/g} \) (latent heat of vaporization)

Calculating \( Q_4 \):
\[ Q_4 = 1 \, \text{g} \cdot 536 \, \text{cal/g} = 536 \, \text{cal} \]

### Total Heat Required
Now we sum all the heat quantities:
\[ Q_{total} = Q_1 + Q_2 + Q_3 + Q_4 \]
\[ Q_{total} = 5 \, \text{cal} + 80 \, \text{cal} + 100 \, \text{cal} + 536 \, \text{cal} \]
\[ Q_{total} = 721 \, \text{cal} \]

Thus, the total amount of heat required to convert 1 gram of ice at -10°C to steam at 100°C is **721 calories**.