1 g of a steam at `100^(@)C` melt how much ice at `0^(@)C`? (Length heat of ice `= 80 cal//gm` and latent heat of steam `= 540 cal//gm`)

Options

A. `1 gm`
B. `2 gm`
C. `4 gm`
D. `8 gm`

Explanation

To solve the problem of how much ice can be melted by 1 gram of steam at 100°C, we will follow these steps:

### Step 1: Calculate the heat released by steam when it condenses to water at 100°C.
The heat released when steam condenses to water is given by the formula:
\[ Q_1 = m \cdot L_s \]
where:
- \( m = 1 \, \text{g} \) (mass of steam)
- \( L_s = 540 \, \text{cal/g} \) (latent heat of steam)

Calculating \( Q_1 \):
\[ Q_1 = 1 \, \text{g} \cdot 540 \, \text{cal/g} = 540 \, \text{cal} \]

### Step 2: Calculate the heat released by water when it cools from 100°C to 0°C.
The heat released when water cools is given by:
\[ Q_2 = m \cdot c \cdot \Delta T \]
where:
- \( c = 1 \, \text{cal/g°C} \) (specific heat capacity of water)
- \( \Delta T = 100°C - 0°C = 100°C \)

Calculating \( Q_2 \):
\[ Q_2 = 1 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 100°C = 100 \, \text{cal} \]

### Step 3: Calculate the total heat released by the steam as it condenses and cools.
The total heat released \( Q \) is the sum of \( Q_1 \) and \( Q_2 \):
\[ Q = Q_1 + Q_2 = 540 \, \text{cal} + 100 \, \text{cal} = 640 \, \text{cal} \]

### Step 4: Calculate the mass of ice that can be melted using the heat released.
The heat required to melt ice is given by:
\[ Q = m \cdot L_i \]
where:
- \( L_i = 80 \, \text{cal/g} \) (latent heat of ice)
- Let \( m \) be the mass of ice melted.

Rearranging the formula gives:
\[ m = \frac{Q}{L_i} = \frac{640 \, \text{cal}}{80 \, \text{cal/g}} = 8 \, \text{g} \]

### Final Answer:
The mass of ice that can be melted by 1 g of steam at 100°C is **8 grams**.

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