One kilogram of steam at `100^@ `C can melt how much ice at `0^@ `C?

Options

A. 8.0 kg
B. `8/54 kg `
C. `54 /8` kg
D. 540 kg

Explanation

To solve the problem of how much ice at \(0^\circ C\) can be melted by \(1 \, \text{kg}\) of steam at \(100^\circ C\), we need to consider the heat exchanges involved in the phase changes and temperature changes. Here’s a step-by-step solution:

### Step 1: Calculate the heat released by steam when it condenses to water at \(100^\circ C\)

When \(1 \, \text{kg}\) of steam condenses to water, it releases heat equal to the latent heat of vaporization. The latent heat of vaporization of water is approximately \(540 \, \text{cal/g}\).

\[ \Delta Q_1 = m \cdot L_v = 1000 \, \text{g} \cdot 540 \, \text{cal/g} = 540000 \, \text{cal} \]

### Step 2: Calculate the heat released by water when it cools from \(100^\circ C\) to \(0^\circ C\)

Next, the \(100^\circ C\) water will cool down to \(0^\circ C\). The specific heat of water is \(1 \, \text{cal/g}^\circ C\).

\[ \Delta Q_2 = m \cdot S \cdot \Delta T = 1000 \, \text{g} \cdot 1 \, \text{cal/g}^\circ C \cdot (100 - 0)^\circ C = 1000 \cdot 100 = 100000 \, \text{cal} \]

### Step 3: Total heat released by the steam

The total heat released by the steam as it condenses and cools can be calculated by adding \(\Delta Q_1\) and \(\Delta Q_2\):

\[ \Delta Q_{total} = \Delta Q_1 + \Delta Q_2 = 540000 \, \text{cal} + 100000 \, \text{cal} = 640000 \, \text{cal} \]

### Step 4: Calculate the heat required to melt ice at \(0^\circ C\)

To melt ice, we need to use the latent heat of fusion. The latent heat of fusion of ice is approximately \(80 \, \text{cal/g}\).

Let \(x\) be the mass of ice melted (in grams). The heat required to melt \(x\) grams of ice is:

\[ \Delta Q_3 = x \cdot L_f = x \cdot 80 \, \text{cal/g} \]

### Step 5: Set the heat released equal to the heat gained

Since the heat released by the steam is equal to the heat gained by the ice, we can set the equations equal:

\[ 640000 \, \text{cal} = x \cdot 80 \, \text{cal/g} \]

### Step 6: Solve for \(x\)

Now, we can solve for \(x\):

\[ x = \frac{640000 \, \text{cal}}{80 \, \text{cal/g}} = 8000 \, \text{g} \]

### Step 7: Convert grams to kilograms

To convert grams to kilograms:

\[ x = 8000 \, \text{g} = 8 \, \text{kg} \]

### Conclusion

Thus, \(1 \, \text{kg}\) of steam at \(100^\circ C\) can melt \(8 \, \text{kg}\) of ice at \(0^\circ C\).

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