How much heat is required to convert 8.0 g of ice at `-15^@` to steam at `100^@`? (Given, `c_(ice) = 0.53 cal//g-^@C, L_f = 80 cal//g and L_v = 539 cal//g, `
and `c_(water) = 1 cal//g-^@C)` .

Explanation

To solve the problem of how much heat is required to convert 8.0 g of ice at -15°C to steam at 100°C, we will break down the process into four steps:

1. **Heating the Ice from -15°C to 0°C**:
We need to calculate the heat required to raise the temperature of the ice from -15°C to 0°C using the specific heat capacity of ice.

\[ Q_1 = m \cdot c_{\text{ice}} \cdot \Delta T \]

Where:
- \( m = 8.0 \, \text{g} \)
- \( c_{\text{ice}} = 0.53 \, \text{cal/g°C} \)
- \( \Delta T = 0 - (-15) = 15 \, \text{°C} \)

Plugging in the values:

\[ Q_1 = 8.0 \, \text{g} \cdot 0.53 \, \text{cal/g°C} \cdot 15 \, \text{°C} = 63.6 \, \text{cal} \]

2. **Melting the Ice at 0°C**:
Next, we calculate the heat required to convert the ice at 0°C to water at 0°C using the latent heat of fusion.

\[ Q_2 = m \cdot L_f \]

Where:
- \( L_f = 80 \, \text{cal/g} \)

Plugging in the values:

\[ Q_2 = 8.0 \, \text{g} \cdot 80 \, \text{cal/g} = 640 \, \text{cal} \]

3. **Heating the Water from 0°C to 100°C**:
Now, we calculate the heat required to raise the temperature of the water from 0°C to 100°C.

\[ Q_3 = m \cdot c_{\text{water}} \cdot \Delta T \]

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

Plugging in the values:

\[ Q_3 = 8.0 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 100 \, \text{°C} = 800 \, \text{cal} \]

4. **Vaporizing the Water at 100°C**:
Finally, we calculate the heat required to convert the water at 100°C to steam at 100°C using the latent heat of vaporization.

\[ Q_4 = m \cdot L_v \]

Where:
- \( L_v = 539 \, \text{cal/g} \)

Plugging in the values:

\[ Q_4 = 8.0 \, \text{g} \cdot 539 \, \text{cal/g} = 4312 \, \text{cal} \]

5. **Total Heat Required**:
Now, we sum up all the heat quantities calculated:

\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 + Q_4 \]

\[ Q_{\text{total}} = 63.6 \, \text{cal} + 640 \, \text{cal} + 800 \, \text{cal} + 4312 \, \text{cal} = 5815.6 \, \text{cal} \]

Therefore, the total heat required to convert 8.0 g of ice at -15°C to steam at 100°C is **5815.6 calories**.