When 0.15 kg at 0°C is mixed with 0.30 kg of water at 50°C in a container, the resulting temperature is 6.7°C. Calculate the heat of fusion of ice.
(S_water = 4186 J.kg⁻¹.K⁻¹)

To solve the problem, we can use the principle of conservation of energy, which states that the heat lost by the warm water will be equal to the heat gained by the cold water (ice) when they reach thermal equilibrium. Here, we need to calculate the heat of fusion of ice given the resulting temperature.

### Given:
- Mass of ice (cold water): \( m_{\text{ice}} = 0.15 \, \text{kg} \)
- Initial temperature of ice: \( T_{\text{ice, initial}} = 0 \, \text{°C} \)
- Mass of warm water: \( m_{\text{water}} = 0.30 \, \text{kg} \)
- Initial temperature of warm water: \( T_{\text{water, initial}} = 50 \, \text{°C} \)
- Final equilibrium temperature: \( T_{\text{final}} = 6.7 \, \text{°C} \)
- Specific heat capacity of water: \( S_{\text{water}} = 4186 \, \text{J/(kg·K)} \)

### Steps to calculate the heat of fusion of ice:

1. **Calculate the heat lost by warm water**:
The heat lost by warm water can be calculated using the formula:
\[
Q_{\text{lost}} = m_{\text{water}} \cdot S_{\text{water}} \cdot (T_{\text{water, initial}} - T_{\text{final}})
\]
Plugging in the values:
\[
Q_{\text{lost}} = 0.30 \, \text{kg} \cdot 4186 \, \text{J/(kg·K)} \cdot (50 - 6.7) \, \text{K}
\]

2. **Calculate the heat gained by the ice**:
The heat gained by the ice consists of two parts: the heat absorbed to melt the ice and the heat absorbed to raise the temperature of the resulting water.
- Heat gained to melt the ice:
\[
Q_{\text{melt}} = m_{\text{ice}} \cdot L_f
\]
where \( L_f \) is the heat of fusion of ice (unknown).

- Heat gained to raise the temperature of the melted ice (now water) to the final temperature:
\[
Q_{\text{raise}} = m_{\text{ice}} \cdot S_{\text{water}} \cdot (T_{\text{final}} - T_{\text{ice, initial}})
\]

3. **Set the heat lost equal to heat gained**:
According to the conservation of energy:
\[
Q_{\text{lost}} = Q_{\text{melt}} + Q_{\text{raise}}
\]
Substituting the equations we derived:
\[
Q_{\text{lost}} = m_{\text{ice}} \cdot L_f + m_{\text{ice}} \cdot S_{\text{water}} \cdot (T_{\text{final}} - T_{\text{ice, initial}})
\]

4. **Plugging in values and solving for \( L_f \)**:
Let's calculate \( Q_{\text{lost}} \) and \( Q_{\text{raise}} \).

### Calculations:

1. **Calculate \( Q_{\text{lost}} \)**:
\[
Q_{\text{lost}} = 0.30 \, \text{kg} \cdot 4186 \, \text{J/(kg·K)} \cdot (50 - 6.7) \, \text{K}
\]
\[
= 0.30 \cdot 4186 \cdot 43.3
\]
\[
= 0.30 \cdot 181,496.8 \, \text{J}
\]
\[
= 54,449.04 \, \text{J}
\]

2. **Calculate \( Q_{\text{raise}} \)**:
\[
Q_{\text{raise}} = 0.15 \, \text{kg} \cdot 4186 \, \text{J/(kg·K)} \cdot (6.7 - 0) \, \text{K}
\]
\[
= 0.15 \cdot 4186 \cdot 6.7
\]
\[
= 0.15 \cdot 28,091.2 \, \text{J}
\]
\[
= 4,213.68 \, \text{J}
\]

3. **Set up the equation**:
\[
54,449.04 \, \text{J} = 0.15 \, \text{kg} \cdot L_f + 4,213.68 \, \text{J}
\]
Subtract \( Q_{\text{raise}} \) from both sides:
\[
54,449.04 \, \text{J} - 4,213.68 \, \text{J} = 0.15 \, \text{kg} \cdot L_f
\]
\[
50,235.36 \, \text{J} = 0.15 \, \text{kg} \cdot L_f
\]

4. **Solve for \( L_f \)**:
\[
L_f = \frac{50,235.36 \, \text{J}}{0.15 \, \text{kg}}
\]
\[
= 334902.4 \, \text{J/kg}
\]
\[
\approx 334.9 \, \text{kJ/kg}
\]

### Final Answer:
The heat of fusion of ice \( L_f \approx 334.9 \, \text{kJ/kg} \).