In an energy recycling process, X g of steam at `100^(@)C` becomes water at `100^(@)C` which converts Y g of ice at `0^(@)C` into water at `100^(@)C`. The ratio of `X/Y` will be (specific heat of water `="4200" J kg"^(-1)K,` specific latent heat of fusion `=3.36xx10^(5)"J kg"^(-2)`, specific latent heat of vaporization `=22.68xx10^(6)" J kg"^(-1)`)

Options

A. `(1)/(3)`
B. `(2)/(3)`
C. 3
D. 2

Answer

To solve the problem, we need to analyze the energy transfer involved in converting steam to water and then using that energy to convert ice to water.

### Step 1: Calculate the heat released by steam
When \( X \) grams of steam at \( 100^\circ C \) condenses into water at \( 100^\circ C \), it releases heat. The heat released \( Q_1 \) can be calculated using the formula:

\[ Q_1 = m \cdot L_v \]

Where:
- \( m = X \times 10^{-3} \) kg (since we need to convert grams to kg)
- \( L_v = 22.68 \times 10^6 \, \text{J/kg} \) (latent heat of vaporization)

Thus,

\[ Q_1 = (X \times 10^{-3}) \cdot (22.68 \times 10^6) \]

### Step 2: Calculate the heat required to convert ice to water
The heat required \( Q_2 \) to convert \( Y \) grams of ice at \( 0^\circ C \) to water at \( 100^\circ C \) consists of two parts:
1. The heat required to melt the ice into water at \( 0^\circ C \).
2. The heat required to raise the temperature of the water from \( 0^\circ C \) to \( 100^\circ C \).

The heat required to melt the ice is given by:

\[ Q_{melt} = m \cdot L_f \]

Where:
- \( m = Y \times 10^{-3} \) kg
- \( L_f = 3.36 \times 10^5 \, \text{J/kg} \) (latent heat of fusion)

Thus,

\[ Q_{melt} = (Y \times 10^{-3}) \cdot (3.36 \times 10^5) \]

The heat required to raise the temperature of the water is given by:

\[ Q_{heat} = m \cdot c \cdot \Delta T \]

Where:
- \( c = 4200 \, \text{J/kg} \cdot \text{K} \) (specific heat of water)
- \( \Delta T = 100^\circ C - 0^\circ C = 100 \, \text{K} \)

Thus,

\[ Q_{heat} = (Y \times 10^{-3}) \cdot (4200) \cdot (100) \]

Combining both parts, the total heat required \( Q_2 \) is:

\[ Q_2 = Q_{melt} + Q_{heat} \]

\[ Q_2 = (Y \times 10^{-3}) \cdot (3.36 \times 10^5) + (Y \times 10^{-3}) \cdot (4200 \cdot 100) \]

\[ Q_2 = (Y \times 10^{-3}) \cdot (3.36 \times 10^5 + 420000) \]

### Step 3: Set the heat released equal to the heat required
Since the heat released by the steam equals the heat required to convert the ice to water, we can set \( Q_1 = Q_2 \):

\[ (X \times 10^{-3}) \cdot (22.68 \times 10^6) = (Y \times 10^{-3}) \cdot (3.36 \times 10^5 + 420000) \]

### Step 4: Simplify the equation
Cancel \( 10^{-3} \) from both sides:

\[ X \cdot (22.68 \times 10^6) = Y \cdot (3.36 \times 10^5 + 420000) \]

### Step 5: Solve for the ratio \( \frac{X}{Y} \)
Rearranging gives:

\[ \frac{X}{Y} = \frac{3.36 \times 10^5 + 420000}{22.68 \times 10^6} \]

Calculating the numerator:

\[ 3.36 \times 10^5 + 420000 = 3.36 \times 10^5 + 4.2 \times 10^5 = 7.56 \times 10^5 \]

Now substituting back:

\[ \frac{X}{Y} = \frac{7.56 \times 10^5}{22.68 \times 10^6} = \frac{7.56}{22.68} \times 10^{-1} = \frac{1}{3} \]

### Final Result
Thus, the ratio \( \frac{X}{Y} \) is:

\[ \frac{X}{Y} = \frac{1}{3} \]