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 of mixing 15 g of ice at 0°C with 10 g of water at 40°C, we need to determine the final temperature of the mixture and the masses of water and ice in the mixture. Here is a step-by-step solution:

### Step 1: Identify the heat transfer
When the ice is mixed with the warm water, heat will flow from the water to the ice. The water will lose heat as it cools down, and the ice will gain heat as it melts.

### Step 2: Calculate the heat lost by the water
The heat lost by the water can be calculated using the formula:
\[ Q = m \cdot C_w \cdot \Delta T \]
where:
- \( m \) = mass of water = 10 g
- \( C_w \) = specific heat capacity of water = 1 cal/g°C
- \( \Delta T \) = change in temperature = (40°C - final temperature)

Let’s assume the final temperature is \( T_f \). The heat lost by the water is:
\[ Q_{water} = 10 \cdot 1 \cdot (40 - T_f) \]

### Step 3: Calculate the heat gained by the ice
The heat gained by the ice will be used to melt it. The heat required to melt ice can be calculated using the formula:
\[ Q = m \cdot L_f \]
where:
- \( m \) = mass of ice = 15 g
- \( L_f \) = latent heat of fusion of ice = 80 cal/g

The heat gained by the ice is:
\[ Q_{ice} = 15 \cdot 80 = 1200 \text{ cal} \]

### Step 4: Set up the equation for heat balance
Since the heat lost by the water will be equal to the heat gained by the ice (assuming no heat loss to the surroundings), we can set up the equation:
\[ 10 \cdot (40 - T_f) = 1200 \]

### Step 5: Solve for the final temperature \( T_f \)
Expanding the equation:
\[ 400 - 10T_f = 1200 \]
Rearranging gives:
\[ 10T_f = 400 - 1200 \]
\[ 10T_f = -800 \]
\[ T_f = -80 \text{ °C} \]

Since this temperature is not physically possible, it indicates that not all the ice melts.

### Step 6: Determine how much ice melts
We need to find out how much ice can melt with the heat available from the water:
\[ Q_{water} = 10 \cdot (40 - 0) = 400 \text{ cal} \]
This means the water can only provide 400 cal to melt the ice.

Let \( m_1 \) be the mass of ice that melts:
\[ m_1 \cdot L_f = 400 \]
\[ m_1 \cdot 80 = 400 \]
\[ m_1 = \frac{400}{80} = 5 \text{ g} \]

### Step 7: Calculate the remaining masses
- Mass of ice remaining = 15 g - 5 g = 10 g
- Mass of water = 10 g (initial water) + 5 g (melted ice) = 15 g

### Step 8: Final temperature of the mixture
After mixing, the final temperature of the mixture will be 0°C since there is still ice present and the mixture will be in equilibrium at the melting point of ice.

### Final Results:
- Final temperature of the mixture: **0°C**
- Mass of water in the mixture: **15 g**
- Mass of ice in the mixture: **10 g**

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