120 g of ice at `0^(@)C` is mixed with 100 g of water at `80^(@)C`. Latent heat of fusion is 80 cal/g and specific heat of water is 1 cal/`g-.^(@)C`. The final temperature of the mixture is

Options

A. `0^(@)C`
B. `40^(@)C`
C. `20^(@)C`
D. `10^(@)C`

Answer

To solve the problem of finding the final temperature when 120 g of ice at 0°C is mixed with 100 g of water at 80°C, we can follow these steps:

### Step 1: Calculate the heat required to melt the ice
The heat required to convert ice at 0°C to water at 0°C can be calculated using the formula:
\[ Q = m \cdot L_f \]
Where:
- \( m = 120 \, \text{g} \) (mass of ice)
- \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion)

Calculating:
\[ Q = 120 \, \text{g} \cdot 80 \, \text{cal/g} = 9600 \, \text{cal} \]

### Step 2: Calculate the heat lost by the warm water
Next, we need to calculate the heat lost by the 100 g of water as it cools from 80°C to 0°C. The heat lost can be calculated using the formula:
\[ Q = m \cdot c \cdot \Delta T \]
Where:
- \( m = 100 \, \text{g} \) (mass of water)
- \( c = 1 \, \text{cal/g°C} \) (specific heat of water)
- \( \Delta T = 80°C - 0°C = 80°C \)

Calculating:
\[ Q = 100 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 80°C = 8000 \, \text{cal} \]

### Step 3: Compare heat gained and lost
Now we compare the heat gained by the ice and the heat lost by the water:
- Heat gained by ice: \( 9600 \, \text{cal} \)
- Heat lost by water: \( 8000 \, \text{cal} \)

Since the heat lost by the water (8000 cal) is less than the heat required to melt the ice (9600 cal), not all the ice will melt.

### Step 4: Determine the final state
Since the heat lost by the water is not sufficient to melt all the ice, the final temperature of the mixture will be 0°C, with some ice remaining.

### Final Answer
The final temperature of the mixture is **0°C**.

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