Calcuate the heat required to convert `0.6` kg of ice at - `20^(@)C`, kept in a calorimeter to steam at `100^(@)C` at atmospheric pressure. Given the specific heat capacity of ice`=2100 j kg^(-1) K^(-1)`, specific heat capacity of water `=4186 j kg^(-1) K^(-1)` latent heat ice `=3.35xx10^(5) j kg^(-1)` and latent heat of steam `=2.256xx10^(6)j kg^(-1)`

Explanation

To calculate the total heat required to convert 0.6 kg of ice at -20°C to steam at 100°C, we need to break down the process into several steps:

### Step 1: Heating the Ice from -20°C to 0°C
We will use the formula:
\[ Q_1 = m \cdot c_{ice} \cdot \Delta T \]
Where:
- \( m = 0.6 \, \text{kg} \) (mass of ice)
- \( c_{ice} = 2100 \, \text{J/kg/K} \) (specific heat capacity of ice)
- \( \Delta T = 0 - (-20) = 20 \, \text{K} \) (temperature change)

Calculating \( Q_1 \):
\[ Q_1 = 0.6 \, \text{kg} \cdot 2100 \, \text{J/kg/K} \cdot 20 \, \text{K} \]
\[ Q_1 = 0.6 \cdot 2100 \cdot 20 = 25200 \, \text{J} \]

### Step 2: Melting the Ice at 0°C to Water at 0°C
We will use the formula:
\[ Q_2 = m \cdot L_{ice} \]
Where:
- \( L_{ice} = 3.35 \times 10^5 \, \text{J/kg} \) (latent heat of fusion)

Calculating \( Q_2 \):
\[ Q_2 = 0.6 \, \text{kg} \cdot 3.35 \times 10^5 \, \text{J/kg} \]
\[ Q_2 = 0.6 \cdot 335000 = 201000 \, \text{J} \]

### Step 3: Heating the Water from 0°C to 100°C
We will use the formula:
\[ Q_3 = m \cdot c_{water} \cdot \Delta T \]
Where:
- \( c_{water} = 4186 \, \text{J/kg/K} \) (specific heat capacity of water)
- \( \Delta T = 100 - 0 = 100 \, \text{K} \) (temperature change)

Calculating \( Q_3 \):
\[ Q_3 = 0.6 \, \text{kg} \cdot 4186 \, \text{J/kg/K} \cdot 100 \, \text{K} \]
\[ Q_3 = 0.6 \cdot 4186 \cdot 100 = 251160 \, \text{J} \]

### Step 4: Converting Water at 100°C to Steam at 100°C
We will use the formula:
\[ Q_4 = m \cdot L_{steam} \]
Where:
- \( L_{steam} = 2.256 \times 10^6 \, \text{J/kg} \) (latent heat of vaporization)

Calculating \( Q_4 \):
\[ Q_4 = 0.6 \, \text{kg} \cdot 2.256 \times 10^6 \, \text{J/kg} \]
\[ Q_4 = 0.6 \cdot 2256000 = 1353600 \, \text{J} \]

### Step 5: Total Heat Required
Now, we sum all the heat calculated in the previous steps:
\[ Q_{total} = Q_1 + Q_2 + Q_3 + Q_4 \]
\[ Q_{total} = 25200 + 201000 + 251160 + 1353600 \]
\[ Q_{total} = 1830960 \, \text{J} \]

Thus, the total heat required to convert 0.6 kg of ice at -20°C to steam at 100°C is **1830960 J**.