50 gram of ice at `0^(@)`C is mixed with 50 gram of water at `60^(@)C` , final temperature of mixture will be :-

Options

A. `0^(@)C`
B. `40^(@)C`
C. `10^(@)C`
D. `15^(@)C`

Explanation

To solve the problem of mixing 50 grams of ice at 0°C with 50 grams of water at 60°C, we need to determine the final temperature of the mixture. Here’s a step-by-step solution:

### Step 1: Calculate the heat lost by the water
The water at 60°C will lose heat as it cools down to 0°C. The amount of heat (Q) lost by the water can be calculated using the formula:

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

Where:
- \(m\) = mass of water = 50 g
- \(C\) = specific heat capacity of water = 1 cal/g°C
- \(\Delta T\) = change in temperature = \(60°C - 0°C = 60°C\)

Substituting the values:

\[ Q = 50 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 60 \, \text{°C} = 3000 \, \text{cal} \]

### Step 2: Calculate the heat required to melt the ice
The heat required to melt the ice can be calculated using the formula:

\[ Q = m_0 \cdot L \]

Where:
- \(m_0\) = mass of ice melted (unknown)
- \(L\) = latent heat of fusion of ice = 80 cal/g

Setting the heat lost by the water equal to the heat gained by the ice:

\[ 3000 \, \text{cal} = m_0 \cdot 80 \, \text{cal/g} \]

Solving for \(m_0\):

\[ m_0 = \frac{3000 \, \text{cal}}{80 \, \text{cal/g}} = 37.5 \, \text{g} \]

### Step 3: Determine the remaining ice and water
Initially, we had 50 g of ice. After melting, the amount of ice remaining is:

\[ \text{Remaining ice} = 50 \, \text{g} - 37.5 \, \text{g} = 12.5 \, \text{g} \]

The total amount of water in the system after mixing is:

\[ \text{Total water} = 50 \, \text{g} + 37.5 \, \text{g} = 87.5 \, \text{g} \]

### Step 4: Determine the final temperature
Since the water cooled down to 0°C and some of the ice melted, but not all of it, the final temperature of the mixture will remain at 0°C. The remaining ice and the melted water will coexist at this temperature.

### Conclusion
The final temperature of the mixture is:

\[ \text{Final Temperature} = 0°C \]