If 10 g of ice at `0^@C` mixes with 10 g of water at `10^@C`, then the final temperature t is given by

Options

A. (10 x 80) + 10 (t-0) =10 (10-t)
B. 10 x 80 = 10 (10-t) + 10 (t-0)
C. `t=5^@` C
D. `t=0^@C`

Answer

To solve the problem of mixing 10 g of ice at 0°C with 10 g of water at 10°C, we need to analyze the heat transfer between the two substances.

### Step-by-Step Solution:

1. **Identify the Heat Required to Melt Ice**:
The latent heat of fusion of ice is given as 80 calories per gram. To melt 10 g of ice, the heat required (Q1) can be calculated as:
\[ Q_1 = \text{mass of ice} \times \text{latent heat} = 10 \, \text{g} \times 80 \, \text{cal/g} = 800 \, \text{calories} \]

2. **Identify the Heat Released by Water**:
The specific heat of water is 1 calorie per gram per degree Celsius. The heat released (Q2) by 10 g of water when it cools from 10°C to 0°C can be calculated as:
\[ Q_2 = \text{mass of water} \times \text{specific heat} \times \Delta T = 10 \, \text{g} \times 1 \, \text{cal/g°C} \times (10 - 0) \, \text{°C} = 100 \, \text{calories} \]

3. **Compare Heat Required and Heat Released**:
We see that the heat required to melt all the ice (800 calories) is greater than the heat released by the water (100 calories). This means that not all the ice will melt.

4. **Calculate the Amount of Ice that Melts**:
The heat released by the water (100 calories) will be used to melt a portion of the ice. We can set up the equation:
\[ Q_2 = m \times L \]
where \( m \) is the mass of ice melted and \( L \) is the latent heat of fusion (80 calories/g). Rearranging gives:
\[ m = \frac{Q_2}{L} = \frac{100 \, \text{cal}}{80 \, \text{cal/g}} = 1.25 \, \text{g} \]

5. **Determine the Final State**:
After melting 1.25 g of ice, we will have:
- Ice remaining: \( 10 \, \text{g} - 1.25 \, \text{g} = 8.75 \, \text{g} \)
- Water from melted ice: \( 1.25 \, \text{g} \)
- Original water: \( 10 \, \text{g} \)

The total amount of water now is:
\[ 1.25 \, \text{g} + 10 \, \text{g} = 11.25 \, \text{g} \]
The final temperature of the mixture will be 0°C, as there is still ice present.

### Final Answer:
The final temperature \( t \) of the mixture is \( 0°C \).