15 g ice at `0^@C` is mixed with 10 g water at `40^@C`. Find the temperature of mixture. Also, find mass of water and ice in the mixture.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the heat lost by the water
We need to find the heat lost by the 10 g of water when it cools from 40 °C to 0 °C. The formula for heat (Q) is:

\[ Q = m \cdot C \cdot \Delta T \]

Where:
- \( m \) = mass of water (10 g)
- \( C \) = specific heat capacity of water (1 cal/g°C)
- \( \Delta T \) = change in temperature (40 °C - 0 °C = 40 °C)

Substituting the values:

\[ Q = 10 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (40 - 0) \, \text{°C} \]
\[ Q = 10 \cdot 1 \cdot 40 = 400 \, \text{calories} \]

### Step 2: Calculate the mass of ice that melts
Now, we will calculate how much ice can melt using the heat absorbed. The latent heat of fusion of ice is given as 80 cal/g.

Let \( M_0 \) be the mass of ice that melts:

\[ M_0 = \frac{Q}{L} \]

Where:
- \( Q \) = heat absorbed (400 calories)
- \( L \) = latent heat of fusion (80 cal/g)

Substituting the values:

\[ M_0 = \frac{400 \, \text{cal}}{80 \, \text{cal/g}} = 5 \, \text{g} \]

### Step 3: Determine the remaining mass of ice
Initially, we had 15 g of ice. Since 5 g of ice has melted, the remaining mass of ice is:

\[ M_{\text{ice remaining}} = 15 \, \text{g} - M_0 \]
\[ M_{\text{ice remaining}} = 15 \, \text{g} - 5 \, \text{g} = 10 \, \text{g} \]

### Step 4: Calculate the total mass of water in the mixture
The total mass of water in the mixture consists of the original 10 g of water plus the 5 g of melted ice:

\[ M_{\text{water}} = 10 \, \text{g} + 5 \, \text{g} = 15 \, \text{g} \]

### Step 5: Determine the final temperature of the mixture
Since the water cools down to 0 °C and the remaining ice is also at 0 °C, the final temperature of the mixture is:

\[ T_{\text{final}} = 0 \, \text{°C} \]

### Final Answers:
- Temperature of the mixture: **0 °C**
- Mass of water in the mixture: **15 g**
- Mass of ice remaining in the mixture: **10 g**