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_4gtQ_3gtQ_2gtQ_1`
B. `Q_4gtQ_3gtQ_1gtQ_2`
C. `Q_4gtQ_2gtQ_3gtQ_1`
D. `Q_4gtQ_2gtQ_1gtQ_3`

Explanation

To solve the problem of calculating the heat required to change 1 kg of ice at -20°C into steam, we will break down the process into four steps, each corresponding to the heat quantities \( Q_1 \), \( Q_2 \), \( Q_3 \), and \( Q_4 \).

### Step 1: Calculate \( Q_1 \) (Heating ice from -20°C to 0°C)
The heat required to raise the temperature of ice from -20°C to 0°C can be calculated using the formula:
\[ Q_1 = m \cdot c_{ice} \cdot \Delta T \]
Where:
- \( m = 1 \, \text{kg} \) (mass of ice)
- \( c_{ice} = 0.5 \, \text{cal/g°C} = 500 \, \text{cal/kg°C} \) (specific heat of ice)
- \( \Delta T = 0 - (-20) = 20 \, \text{°C} \)

Substituting the values:
\[ Q_1 = 1 \, \text{kg} \cdot 500 \, \text{cal/kg°C} \cdot 20 \, \text{°C} = 10000 \, \text{cal} \]

### Step 2: Calculate \( Q_2 \) (Melting the ice)
The heat required to melt the ice at 0°C can be calculated using the formula:
\[ Q_2 = m \cdot L_f \]
Where:
- \( L_f = 80 \, \text{cal/kg} \) (latent heat of fusion)

Substituting the values:
\[ Q_2 = 1 \, \text{kg} \cdot 80 \, \text{cal/kg} = 80 \, \text{cal} \]

### Step 3: Calculate \( Q_3 \) (Heating water from 0°C to 100°C)
The heat required to raise the temperature of water from 0°C to 100°C can be calculated using the formula:
\[ Q_3 = m \cdot c_{water} \cdot \Delta T \]
Where:
- \( c_{water} = 1 \, \text{cal/g°C} = 1000 \, \text{cal/kg°C} \) (specific heat of water)
- \( \Delta T = 100 - 0 = 100 \, \text{°C} \)

Substituting the values:
\[ Q_3 = 1 \, \text{kg} \cdot 1000 \, \text{cal/kg°C} \cdot 100 \, \text{°C} = 100000 \, \text{cal} \]

### Step 4: Calculate \( Q_4 \) (Vaporizing the water)
The heat required to vaporize the water at 100°C can be calculated using the formula:
\[ Q_4 = m \cdot L_v \]
Where:
- \( L_v = 540 \, \text{cal/kg} \) (latent heat of vaporization)

Substituting the values:
\[ Q_4 = 1 \, \text{kg} \cdot 540 \, \text{cal/kg} = 540 \, \text{cal} \]

### Summary of Heat Quantities
Now we can summarize the heat quantities:
- \( Q_1 = 10000 \, \text{cal} \)
- \( Q_2 = 80 \, \text{cal} \)
- \( Q_3 = 100000 \, \text{cal} \)
- \( Q_4 = 540 \, \text{cal} \)

### Final Relation
From the calculations, we can see the relationship between the heat quantities:
\[ Q_4 > Q_3 > Q_2 > Q_1 \]