When a `0.45` kg of ice at `0^(@)C` mixed with `0.9` kg of water at `55^(@)C` in a container, the resulting temperature is `10^(@)C`. Calculate the heat of fusion of ice. Specific heat of water is `4186 j kg^(-1) K^(-1)`.

Explanation

To solve the problem, we need to calculate the heat of fusion of ice when it is mixed with water at a different temperature. Here’s a step-by-step solution:

### Step 1: Calculate the heat lost by the water
The heat lost by the water as it cools from 55°C to 10°C can be calculated using the formula:
\[ Q_1 = m_w \cdot c_w \cdot \Delta T_w \]
Where:
- \( m_w = 0.9 \, \text{kg} \) (mass of water)
- \( c_w = 4186 \, \text{J/kg/K} \) (specific heat of water)
- \( \Delta T_w = (55 - 10) \, \text{°C} = 45 \, \text{°C} \)

Substituting the values:
\[ Q_1 = 0.9 \, \text{kg} \cdot 4186 \, \text{J/kg/K} \cdot 45 \, \text{K} \]
\[ Q_1 = 0.9 \cdot 4186 \cdot 45 \]
\[ Q_1 = 168,168 \, \text{J} \]

### Step 2: Calculate the heat gained by the ice
The heat gained by the ice consists of two parts: the heat required to melt the ice and the heat required to raise the temperature of the resulting water from 0°C to 10°C. This can be expressed as:
\[ Q_2 = m_i \cdot L_f + m_i \cdot c_w \cdot \Delta T_i \]
Where:
- \( m_i = 0.45 \, \text{kg} \) (mass of ice)
- \( L_f \) = latent heat of fusion of ice (unknown)
- \( \Delta T_i = (10 - 0) \, \text{°C} = 10 \, \text{°C} \)

Substituting the known values:
\[ Q_2 = 0.45 \cdot L_f + 0.45 \cdot 4186 \cdot 10 \]
\[ Q_2 = 0.45 \cdot L_f + 18837 \, \text{J} \]

### Step 3: Set the heat lost equal to the heat gained
Since there is no heat loss to the surroundings, we can set \( Q_1 \) equal to \( Q_2 \):
\[ Q_1 = Q_2 \]
\[ 168168 = 0.45 \cdot L_f + 18837 \]

### Step 4: Solve for the latent heat of fusion \( L_f \)
Rearranging the equation to isolate \( L_f \):
\[ 168168 - 18837 = 0.45 \cdot L_f \]
\[ 149331 = 0.45 \cdot L_f \]
\[ L_f = \frac{149331}{0.45} \]
\[ L_f = 331,000 \, \text{J/kg} \]

### Conclusion
The heat of fusion of ice is approximately \( 331,000 \, \text{J/kg} \).

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