The final temperature, when `10 g` of steam at `100^@ C` is passed into an ice block of mass `100 g`
`(L_(steam) = 540 cal//g, L_(ice) = 80 cal//g , S_(water) =1 cal//g^(@) C)` is

Options

A. `0^@C`
B. `15.7^@C`
C. `16.9^@C`
D. `4^@ C`

Explanation

To solve the problem, we need to analyze the heat transfer between the steam and the ice block. The steam will condense into water and cool down, while the ice will absorb heat and melt. We will determine the final temperature based on the energy balance.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of steam, \( m_{steam} = 10 \, \text{g} \)
- Mass of ice, \( m_{ice} = 100 \, \text{g} \)
- Latent heat of steam, \( L_{steam} = 540 \, \text{cal/g} \)
- Latent heat of ice, \( L_{ice} = 80 \, \text{cal/g} \)
- Specific heat of water, \( S_{water} = 1 \, \text{cal/g}^\circ C \)

2. **Calculate the Heat Required to Melt the Ice:**
\[ Q_{melt} = m_{ice} \times L_{ice} = 100 \, \text{g} \times 80 \, \text{cal/g} = 8000 \, \text{cal} \]

3. **Calculate the Heat Released by the Steam:**
- When steam condenses to water at \(100^\circ C\), it releases heat:
\[ Q_{condense} = m_{steam} \times L_{steam} = 10 \, \text{g} \times 540 \, \text{cal/g} = 5400 \, \text{cal} \]
- The steam then cools from \(100^\circ C\) to \(0^\circ C\):
\[ Q_{cool} = m_{steam} \times S_{water} \times \Delta T = 10 \, \text{g} \times 1 \, \text{cal/g}^\circ C \times (100 - 0) = 1000 \, \text{cal} \]
- Total heat released by the steam:
\[ Q_{total\_released} = Q_{condense} + Q_{cool} = 5400 \, \text{cal} + 1000 \, \text{cal} = 6400 \, \text{cal} \]

4. **Compare the Heat Required to Melt the Ice and the Heat Released by the Steam:**
- Heat required to melt the ice: \( Q_{melt} = 8000 \, \text{cal} \)
- Heat released by the steam: \( Q_{total\_released} = 6400 \, \text{cal} \)

5. **Conclusion:**
Since the heat released by the steam (6400 cal) is less than the heat required to melt the ice (8000 cal), the final temperature of the system will be \(0^\circ C\). The steam will not be able to melt all the ice, and thus, the final temperature will remain at \(0^\circ C\).

### Final Answer:
The final temperature when \(10 \, \text{g}\) of steam at \(100^\circ C\) is passed into an ice block of mass \(100 \, \text{g}\) is \(0^\circ C\).

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