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 of the mixture when 120 g of ice at 0°C is mixed with 100 g of water at 80°C, we will 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 water
Next, we calculate the heat lost by the 100 g of water as it cools from 80°C to 0°C:
\[ Q = m \cdot s \cdot \Delta T \]
where:
- \( m = 100 \, \text{g} \) (mass of water)
- \( s = 1 \, \text{cal/g°C} \) (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: Determine the net heat exchange
Now, we find the net heat exchange. The heat required to melt the ice (9600 cal) is greater than the heat lost by the water (8000 cal). Therefore, we calculate the remaining heat after the ice has melted:
\[ Q_{\text{remaining}} = Q_{\text{melted ice}} - Q_{\text{lost by water}} \]
\[ Q_{\text{remaining}} = 9600 \, \text{cal} - 8000 \, \text{cal} = 1600 \, \text{cal} \]

### Step 4: Calculate the final temperature of the mixture
After the ice has melted, we have 220 g of water (120 g from melted ice + 100 g of original water) at 0°C. We need to find the final temperature \( T_f \) when this water absorbs the remaining heat:
\[ Q = m \cdot s \cdot \Delta T \]
where:
- \( m = 220 \, \text{g} \)
- \( s = 1 \, \text{cal/g°C} \)
- \( \Delta T = T_f - 0°C \)

Setting the heat absorbed equal to the remaining heat:
\[ 1600 \, \text{cal} = 220 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (T_f - 0) \]

This simplifies to:
\[ 1600 = 220 T_f \]

### Step 5: Solve for \( T_f \)
Now, we can solve for \( T_f \):
\[ T_f = \frac{1600}{220} = \frac{80}{11} \approx 7.27°C \]

### Conclusion
The final temperature of the mixture is approximately \( 7.27°C \).

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