To solve the problem, we will use the principle of calorimetry, which states that the heat lost by the hot water will be equal to the heat gained by the ice as it melts and warms up.
### Step-by-Step Solution:
1. **Identify the Given Values:**
- Mass of ice, \( m_{ice} = 1.5 \, \text{kg} \)
- Initial temperature of ice, \( T_{ice} = 0^\circ C \)
- Mass of water, \( m_{water} = 2 \, \text{kg} \)
- Initial temperature of water, \( T_{water} = 70^\circ C \)
- Final temperature, \( T_{final} = 5^\circ C \)
- Specific heat of water, \( s_{water} = 4186 \, \text{J/kg/K} \)
2. **Calculate the Heat Lost by Water:**
The heat lost by the water as it cools down from \( 70^\circ C \) to \( 5^\circ C \) can be calculated using the formula:
\[
Q_{lost} = m_{water} \cdot s_{water} \cdot (T_{initial} - T_{final})
\]
Substituting the values:
\[
Q_{lost} = 2 \, \text{kg} \cdot 4186 \, \text{J/kg/K} \cdot (70 - 5)
\]
\[
Q_{lost} = 2 \cdot 4186 \cdot 65
\]
\[
Q_{lost} = 544180 \, \text{J}
\]
3. **Calculate the Heat Gained by 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 melted ice (water) from \( 0^\circ C \) to \( 5^\circ C \).
- Heat required to melt the ice:
\[
Q_{gain, melt} = m_{ice} \cdot L
\]
- Heat required to raise the temperature of the melted ice:
\[
Q_{gain, heat} = m_{ice} \cdot s_{water} \cdot (T_{final} - T_{ice})
\]
Therefore, the total heat gained by the ice is:
\[
Q_{gain} = Q_{gain, melt} + Q_{gain, heat}
\]
Substituting the values:
\[
Q_{gain} = m_{ice} \cdot L + m_{ice} \cdot s_{water} \cdot (5 - 0)
\]
\[
Q_{gain} = 1.5 \cdot L + 1.5 \cdot 4186 \cdot 5
\]
\[
Q_{gain} = 1.5 \cdot L + 1.5 \cdot 20930
\]
\[
Q_{gain} = 1.5 \cdot L + 31395
\]
4. **Set Heat Lost Equal to Heat Gained:**
According to the principle of calorimetry:
\[
Q_{lost} = Q_{gain}
\]
\[
544180 = 1.5 \cdot L + 31395
\]
5. **Solve for \( L \):**
Rearranging the equation:
\[
1.5 \cdot L = 544180 - 31395
\]
\[
1.5 \cdot L = 512785
\]
\[
L = \frac{512785}{1.5}
\]
\[
L = 341857 \, \text{J/kg}
\]
6. **Final Result:**
The heat of fusion of ice is approximately:
\[
L \approx 3.42 \times 10^5 \, \text{J/kg}
\]
