What is the amount of heat required (in calories) to convert 10 g of ice at `-10^(@)C` into steam at `100^(@)C`? Given that latent heat of vaporization of water is `"540 cal g"^(-1)`, latent heat of fusion of ice is `"80 cal g"^(-1)`, the specific heat capacity of water and ice are `"1 cal g"^(-1).^(@)C^(-1)` and `"0.5 cal g"^(-1).^(@)C^(-1)` respectively.

Answer

To find the total amount of heat required to convert 10 g of ice at -10°C into steam at 100°C, we need to consider several steps in the heating and phase change processes. Here’s a step-by-step breakdown of the calculations:

### Step 1: Heating the Ice from -10°C to 0°C
We first need to heat the ice from -10°C to 0°C. The specific heat capacity of ice is given as 0.5 cal/g°C.

\[ Q_1 = m \cdot c_{ice} \cdot \Delta T \]
Where:
- \( m = 10 \, \text{g} \)
- \( c_{ice} = 0.5 \, \text{cal/g°C} \)
- \( \Delta T = 0 - (-10) = 10°C \)

Calculating \( Q_1 \):
\[ Q_1 = 10 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot 10°C = 50 \, \text{cal} \]

### Step 2: Melting the Ice at 0°C to Water
Next, we convert the ice at 0°C to water at 0°C using the latent heat of fusion, which is 80 cal/g.

\[ Q_2 = m \cdot L_f \]
Where:
- \( L_f = 80 \, \text{cal/g} \)

Calculating \( Q_2 \):
\[ Q_2 = 10 \, \text{g} \cdot 80 \, \text{cal/g} = 800 \, \text{cal} \]

### Step 3: Heating the Water from 0°C to 100°C
Now, we need to heat the water from 0°C to 100°C. The specific heat capacity of water is given as 1 cal/g°C.

\[ Q_3 = m \cdot c_{water} \cdot \Delta T \]
Where:
- \( c_{water} = 1 \, \text{cal/g°C} \)
- \( \Delta T = 100 - 0 = 100°C \)

Calculating \( Q_3 \):
\[ Q_3 = 10 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 100°C = 1000 \, \text{cal} \]

### Step 4: Converting Water at 100°C to Steam
Finally, we convert the water at 100°C to steam using the latent heat of vaporization, which is 540 cal/g.

\[ Q_4 = m \cdot L_v \]
Where:
- \( L_v = 540 \, \text{cal/g} \)

Calculating \( Q_4 \):
\[ Q_4 = 10 \, \text{g} \cdot 540 \, \text{cal/g} = 5400 \, \text{cal} \]

### Step 5: Total Heat Required
Now, we sum all the heat quantities calculated in the previous steps to find the total heat required.

\[ Q_{total} = Q_1 + Q_2 + Q_3 + Q_4 \]
Calculating \( Q_{total} \):
\[ Q_{total} = 50 \, \text{cal} + 800 \, \text{cal} + 1000 \, \text{cal} + 5400 \, \text{cal} = 7250 \, \text{cal} \]

### Final Answer
The total amount of heat required to convert 10 g of ice at -10°C into steam at 100°C is **7250 calories**.

---