To find the latent heat of fusion of ice in the given scenario, we will use the principle of conservation of energy. The heat lost by the warm water will be equal to the heat gained by the ice as it melts and warms up.
### Step-by-Step Solution:
**Step 1: Calculate the heat lost by the warm water when 50 g of ice is added.**
- Initial mass of water, \( m_w = 200 \, \text{g} \)
- Initial temperature of water, \( T_{w_i} = 70^\circ C \)
- Final temperature after adding ice, \( T_f = 40^\circ C \)
- Specific heat capacity of water, \( C_w = 1 \, \text{cal/g}^\circ C \)
The heat lost by the water can be calculated using the formula:
\[
Q_w = m_w \cdot C_w \cdot (T_{w_i} - T_f)
\]
Substituting the values:
\[
Q_w = 200 \cdot 1 \cdot (70 - 40) = 200 \cdot 30 = 6000 \, \text{cal}
\]
**Step 2: Calculate the heat gained by the ice.**
- Mass of ice added, \( m_i = 50 \, \text{g} \)
- Latent heat of fusion of ice, \( L_f \) (unknown)
- The ice melts and then warms up to the final temperature \( T_f = 40^\circ C \).
The heat gained by the ice consists of two parts:
1. Heat required to melt the ice:
\[
Q_{melt} = m_i \cdot L_f = 50 \cdot L_f
\]
2. Heat required to raise the temperature of the melted ice (now water) from \( 0^\circ C \) to \( 40^\circ C \):
\[
Q_{warm} = m_i \cdot C_w \cdot (T_f - 0) = 50 \cdot 1 \cdot (40 - 0) = 2000 \, \text{cal}
\]
The total heat gained by the ice is:
\[
Q_i = Q_{melt} + Q_{warm} = 50L_f + 2000
\]
**Step 3: Set the heat lost equal to the heat gained.**
From the conservation of energy:
\[
Q_w = Q_i
\]
Substituting the expressions we found:
\[
6000 = 50L_f + 2000
\]
**Step 4: Solve for \( L_f \).**
Rearranging the equation:
\[
6000 - 2000 = 50L_f
\]
\[
4000 = 50L_f
\]
\[
L_f = \frac{4000}{50} = 80 \, \text{cal/g}
\]
**Step 5: Repeat the process for the second addition of ice (80 g) and find a second equation.**
Now, when 80 g of ice is added and melts, the new total mass of water becomes \( 250 \, \text{g} \) (200 g + 50 g). The final temperature is \( 10^\circ C \).
- New heat lost by the water:
\[
Q_w' = 250 \cdot 1 \cdot (40 - 10) = 250 \cdot 30 = 7500 \, \text{cal}
\]
- Heat gained by the additional ice:
1. Heat required to melt the ice:
\[
Q_{melt}' = 80L_f
\]
2. Heat required to raise the temperature of the melted ice from \( 0^\circ C \) to \( 10^\circ C \):
\[
Q_{warm}' = 80 \cdot 1 \cdot (10 - 0) = 800 \, \text{cal}
\]
Total heat gained by the ice:
\[
Q_i' = Q_{melt}' + Q_{warm}' = 80L_f + 800
\]
Setting the heat lost equal to the heat gained:
\[
7500 = 80L_f + 800
\]
Solving for \( L_f \):
\[
7500 - 800 = 80L_f
\]
\[
6700 = 80L_f
\]
\[
L_f = \frac{6700}{80} = 83.75 \, \text{cal/g}
\]
### Final Calculation:
Now, we have two equations for \( L_f \):
1. From the first case: \( L_f = 80 \, \text{cal/g} \)
2. From the second case: \( L_f = 83.75 \, \text{cal/g} \)
To find an average or a more accurate value, we can average these values or take the most consistent one based on experimental conditions.
### Final Answer:
The latent heat of fusion of ice is approximately \( 80 \, \text{cal/g} \).
