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 calculate the total heat required to convert 8.0 g of ice at -15°C to steam at 100°C, we will follow these steps:

### Step 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 formula:
\[ Q_1 = m \cdot c_{\text{ice}} \cdot \Delta T \]
Where:
- \(m = 8.0 \, \text{g}\) (mass of ice)
- \(c_{\text{ice}} = 0.53 \, \text{cal/g°C}\) (specific heat of ice)
- \(\Delta T = 0 - (-15) = 15 \, \text{°C}\) (temperature change)

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

### Step 2: Melting the Ice at 0°C
Next, we need to calculate the heat required to convert the ice at 0°C to water at 0°C:
\[ Q_2 = m \cdot L_f \]
Where:
- \(L_f = 80 \, \text{cal/g}\) (latent heat of fusion)

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

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

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

### Step 4: Converting Water at 100°C to Steam
Finally, we need to calculate the heat required to convert the water at 100°C to steam at 100°C:
\[ Q_4 = m \cdot L_v \]
Where:
- \(L_v = 539 \, \text{cal/g}\) (latent heat of vaporization)

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

### Step 5: Total Heat Required
Now we can sum up all the heat quantities to find the total heat required:
\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 + Q_4 \]
Calculating \(Q_{\text{total}}\):
\[ Q_{\text{total}} = 63.6 \, \text{cal} + 640 \, \text{cal} + 800 \, \text{cal} + 4312 \, \text{cal} = 5815.6 \, \text{cal} \]

### Final Answer
The total heat required to convert 8.0 g of ice at -15°C to steam at 100°C is approximately **5816 cal**.

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