Heat is required to change 1 kg of ice at `-20^@C` into steam. `Q_1` is the heat needed to warm the ice from `-20^@C` to `0^@C`, `Q_2` is the heat needed to melt the ice, `Q_3` is the heat needed to warm the water from `0^@C` to `100^@C` and `Q_4` is the heat needed to vapourize the water. Then

Options

A. `Q_(4)gtQ_(3)gtQ_(2)gtQ_(1)`
B. `Q_(4)gtQ_(3)gtQ_(1)gtQ_(2)`
C. `Q_(4)gtQ_(2)gtQ_(3)gtQ_(1)`
D. `Q_(4)gtQ_(2)gtQ_(1)gtQ_(3)`

Explanation

To solve the problem of calculating the total heat required to change 1 kg of ice at -20°C into steam, we need to break down the process into four distinct steps, each associated with a specific amount of heat (Q1, Q2, Q3, and Q4).

### Step-by-Step Solution:

1. **Calculate Q1: Heating the Ice from -20°C to 0°C**
- Formula: \( Q_1 = m \cdot s_{ice} \cdot \Delta T \)
- Where:
- \( m = 1 \, \text{kg} = 1000 \, \text{g} \)
- \( s_{ice} = 1 \, \text{cal/g°C} \)
- \( \Delta T = 0 - (-20) = 20 \, \text{°C} \)
- Calculation:
\[ Q_1 = 1000 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 20 \, \text{°C} = 20000 \, \text{cal} \]

2. **Calculate Q2: Melting the Ice at 0°C**
- Formula: \( Q_2 = m \cdot L_f \)
- Where:
- \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion)
- Calculation:
\[ Q_2 = 1000 \, \text{g} \cdot 80 \, \text{cal/g} = 80000 \, \text{cal} \]

3. **Calculate Q3: Heating the Water from 0°C to 100°C**
- Formula: \( Q_3 = m \cdot s_{water} \cdot \Delta T \)
- Where:
- \( s_{water} = 1 \, \text{cal/g°C} \)
- \( \Delta T = 100 - 0 = 100 \, \text{°C} \)
- Calculation:
\[ Q_3 = 1000 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 100 \, \text{°C} = 100000 \, \text{cal} \]

4. **Calculate Q4: Vaporizing the Water at 100°C**
- Formula: \( Q_4 = m \cdot L_v \)
- Where:
- \( L_v = 540 \, \text{cal/g} \) (latent heat of vaporization)
- Calculation:
\[ Q_4 = 1000 \, \text{g} \cdot 540 \, \text{cal/g} = 540000 \, \text{cal} \]

5. **Total Heat Required (Q_total)**
- Formula: \( Q_{total} = Q_1 + Q_2 + Q_3 + Q_4 \)
- Calculation:
\[ Q_{total} = 20000 \, \text{cal} + 80000 \, \text{cal} + 100000 \, \text{cal} + 540000 \, \text{cal} = 740000 \, \text{cal} \]

### Final Answer:
The total heat required to change 1 kg of ice at -20°C into steam is **740000 calories**.