If 10 g of ice is added to 40 g of water at `15^(@)C`, then the temperature of the mixture is (specific heat of water `= 4.2 xx 10^(3) j kg^(-1) K^(-1)`, Latent heat of fusion of ice `= 3.36 xx 10^(5) j kg^(-1)`)

Options

A. `15^(@)C`
B. `12^(@)C`
C. `10^(2)C`
D. `0^(@)C`

Explanation

To solve the problem of finding the final temperature of the mixture when 10 g of ice is added to 40 g of water at 15°C, we will follow these steps:

### Step 1: Identify the given data
- Mass of ice, \( m_{\text{ice}} = 10 \, \text{g} = 0.01 \, \text{kg} \)
- Mass of water, \( m_{\text{water}} = 40 \, \text{g} = 0.04 \, \text{kg} \)
- Initial temperature of water, \( T_{\text{water}} = 15 \, \text{°C} \)
- Initial temperature of ice, \( T_{\text{ice}} = 0 \, \text{°C} \)
- Specific heat of water, \( c_{\text{water}} = 4.2 \times 10^3 \, \text{J/kg/K} \)
- Latent heat of fusion of ice, \( L_f = 3.36 \times 10^5 \, \text{J/kg} \)

### Step 2: Calculate the heat lost by the water as it cools down to 0°C
The heat lost by the water when it cools from 15°C to 0°C can be calculated using the formula:
\[ Q_{\text{lost}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T \]
Where \( \Delta T = T_{\text{initial}} - T_{\text{final}} = 15 - 0 = 15 \, \text{°C} \).

Substituting the values:
\[ Q_{\text{lost}} = 0.04 \, \text{kg} \cdot 4.2 \times 10^3 \, \text{J/kg/K} \cdot 15 \, \text{K} \]
\[ Q_{\text{lost}} = 0.04 \cdot 4.2 \times 10^3 \cdot 15 = 2520 \, \text{J} \]

### Step 3: Calculate the heat required to melt the ice
The heat required to convert ice at 0°C to water at 0°C is given by:
\[ Q_{\text{gained}} = m_{\text{ice}} \cdot L_f \]
Substituting the values:
\[ Q_{\text{gained}} = 0.01 \, \text{kg} \cdot 3.36 \times 10^5 \, \text{J/kg} \]
\[ Q_{\text{gained}} = 0.01 \cdot 3.36 \times 10^5 = 3360 \, \text{J} \]

### Step 4: Compare the heat lost and gained
Now we compare the heat lost by the water and the heat gained by the ice:
- Heat lost by water: \( Q_{\text{lost}} = 2520 \, \text{J} \)
- Heat required to melt the ice: \( Q_{\text{gained}} = 3360 \, \text{J} \)

Since \( Q_{\text{lost}} < Q_{\text{gained}} \), the water does not have enough heat to melt all the ice. Therefore, not all the ice will melt, and the final temperature of the mixture will remain at 0°C.

### Conclusion
The final temperature of the mixture is \( 0 \, \text{°C} \).

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