5 gm of steam at `100^(@)C` is passed into six gm of ice at `0^(@)C`. If the latent heats of steam and ice in cal per gm are 540 and 80 respectively, then the final temperature is -

Options

A. `0^(@)C`
B. `100^(@)C`
C. `50^(@)C`
D. `30^(@)C`

Answer

To solve the problem of finding the final temperature when 5 gm of steam at 100°C is passed into 6 gm of ice at 0°C, we will follow these steps:

### Step 1: Calculate the heat required to convert ice to water at 0°C
The heat required to convert ice to water can be calculated using the formula:
\[ Q = m \cdot L \]
where:
- \( m \) = mass of ice = 6 gm
- \( L \) = latent heat of fusion of ice = 80 cal/gm

Calculating:
\[ Q = 6 \, \text{gm} \times 80 \, \text{cal/gm} = 480 \, \text{cal} \]

### Step 2: Calculate the heat required to raise the temperature of water from 0°C to 100°C
The heat required to raise the temperature of the water can be calculated using the formula:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \( m \) = mass of water = 6 gm
- \( c \) = specific heat of water = 1 cal/gm°C
- \( \Delta T \) = change in temperature = 100°C - 0°C = 100°C

Calculating:
\[ Q = 6 \, \text{gm} \times 1 \, \text{cal/gm°C} \times 100 \, \text{°C} = 600 \, \text{cal} \]

### Step 3: Calculate the total heat required to convert ice at 0°C to water at 100°C
Total heat required:
\[ Q_{\text{total}} = Q_{\text{ice to water}} + Q_{\text{water heating}} \]
\[ Q_{\text{total}} = 480 \, \text{cal} + 600 \, \text{cal} = 1080 \, \text{cal} \]

### Step 4: Calculate the heat released by steam when it condenses to water at 100°C
The heat released by the steam can be calculated using:
\[ Q = m \cdot L \]
where:
- \( m \) = mass of steam = 5 gm
- \( L \) = latent heat of vaporization of steam = 540 cal/gm

Calculating:
\[ Q = 5 \, \text{gm} \times 540 \, \text{cal/gm} = 2700 \, \text{cal} \]

### Step 5: Determine the net heat available after converting ice to water
Net heat available after converting ice to water:
\[ Q_{\text{remaining}} = Q_{\text{steam}} - Q_{\text{total}} \]
\[ Q_{\text{remaining}} = 2700 \, \text{cal} - 1080 \, \text{cal} = 1620 \, \text{cal} \]

### Step 6: Calculate the amount of water that can be vaporized using the remaining heat
The amount of water that can be vaporized can be calculated using:
\[ \text{mass of vapor} = \frac{Q_{\text{remaining}}}{L} \]
where:
- \( L \) = latent heat of vaporization = 540 cal/gm

Calculating:
\[ \text{mass of vapor} = \frac{1620 \, \text{cal}}{540 \, \text{cal/gm}} = 3 \, \text{gm} \]

### Step 7: Determine the final state of the system
The total mass of water after condensation of steam and melting of ice:
- Initial mass of water from steam = 5 gm
- Mass of ice melted = 6 gm
- Mass of vaporized water = 3 gm

Total mass of water:
\[ \text{Total water} = 6 \, \text{gm} + 5 \, \text{gm} - 3 \, \text{gm} = 8 \, \text{gm} \]

### Step 8: Determine the final temperature
Since all the processes occur at 100°C (the steam condenses at this temperature), the final temperature of the system will be:
\[ \text{Final Temperature} = 100°C \]

Thus, the final temperature of the system is **100°C**.

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