If `10 g` of the ice at `0^@C` is mixed with `10g` of water at `100^@ C`, then the final temperature of the mixture will be

Options

A. `5^@C`
B. `10^@ C`
C. `100 K`
D. `0^@ C`

Explanation

To solve the problem of finding the final temperature when `10 g` of ice at `0°C` is mixed with `10 g` of water at `100°C`, we can follow these steps:

### Step 1: Calculate the heat required to melt the ice
To convert ice at `0°C` to water at `0°C`, we need to use the latent heat of fusion. The latent heat of fusion of ice is approximately `80 calories/g`.

For `10 g` of ice:
\[ \text{Heat required to melt ice} = \text{mass} \times \text{latent heat of fusion} = 10 \, \text{g} \times 80 \, \text{cal/g} = 800 \, \text{calories} \]

### Step 2: Calculate the heat released by the water
Next, we need to calculate how much heat the `10 g` of water at `100°C` can release when it cools down to `0°C`. The specific heat of water is `1 calorie/g°C`.

The total heat released when the water cools from `100°C` to `0°C` is:
\[ \text{Heat released by water} = \text{mass} \times \text{specific heat} \times \text{temperature change} = 10 \, \text{g} \times 1 \, \text{cal/g°C} \times (100 - 0) \, °C = 1000 \, \text{calories} \]

### Step 3: Compare the heat required and heat released
Now we compare the heat required to melt the ice and the heat released by the water:
- Heat required to melt the ice: `800 calories`
- Heat released by the water: `1000 calories`

Since the heat released by the water (`1000 calories`) is greater than the heat required to melt the ice (`800 calories`), all the ice will melt, and there will still be some heat left over.

### Step 4: Calculate the remaining heat after melting the ice
After melting the ice, the remaining heat is:
\[ \text{Remaining heat} = \text{Heat released by water} - \text{Heat required to melt ice} = 1000 \, \text{calories} - 800 \, \text{calories} = 200 \, \text{calories} \]

### Step 5: Calculate the final temperature of the mixture
After the ice has melted, we have `10 g` of water from the melted ice at `0°C` and `10 g` of water from the original water at `0°C`. The total mass of the water is:
\[ \text{Total mass} = 10 \, \text{g} + 10 \, \text{g} = 20 \, \text{g} \]

Now we can use the remaining heat to find the final temperature. The specific heat of water is `1 calorie/g°C`, so the temperature rise can be calculated using:
\[ \text{Heat} = \text{mass} \times \text{specific heat} \times \text{temperature change} \]
Let \( T_f \) be the final temperature:
\[ 200 \, \text{calories} = 20 \, \text{g} \times 1 \, \text{cal/g°C} \times (T_f - 0°C) \]
\[ 200 = 20 \times T_f \]
\[ T_f = \frac{200}{20} = 10°C \]

### Final Answer
The final temperature of the mixture is \( 10°C \).

---