Ice at `0^@C` is added to 200 g of water initially at `70^@C` in a vacuum flask. When 50 g of ice has been added and has all melted the temperature of the flask and contents is `40^@C`. When a further 80 g of ice has been added and has all melted the temperature of the whole becomes `10^@C`. Find the latent heat of fusion of ice.

Explanation

To find the latent heat of fusion of ice, we can follow these steps:

### Step 1: Understand the system
We have:
- 200 g of water at 70°C.
- 50 g of ice at 0°C is added, resulting in a temperature of 40°C.
- An additional 80 g of ice is added, resulting in a final temperature of 10°C.

### Step 2: Calculate heat lost by water and flask when 50 g of ice is added
1. **Heat lost by water**:
- Mass of water (m_w) = 200 g
- Initial temperature (T_initial) = 70°C
- Final temperature (T_final) = 40°C
- Change in temperature (ΔT_w) = T_initial - T_final = 70°C - 40°C = 30°C
- Heat lost by water (Q_w) = m_w * s * ΔT_w = 200 g * 1 cal/g°C * 30°C = 6000 cal

2. **Heat lost by the flask**:
- Let the heat capacity of the flask be C.
- Change in temperature (ΔT_f) = 70°C - 40°C = 30°C
- Heat lost by the flask (Q_f) = C * ΔT_f = C * 30°C

3. **Heat gained by the ice**:
- Mass of ice (m_i) = 50 g
- Latent heat of fusion (L_f) = ?
- Heat gained by ice (Q_i) = m_i * L_f + m_i * s * ΔT_i
- Change in temperature for ice (ΔT_i) = 40°C - 0°C = 40°C
- Heat gained by ice (Q_i) = 50 g * L_f + 50 g * 1 cal/g°C * 40°C = 50 L_f + 2000 cal

### Step 3: Set up the equation for the first scenario
From the conservation of energy:
\[ Q_w + Q_f = Q_i \]
Substituting the values:
\[ 6000 + C * 30 = 50 L_f + 2000 \]
Rearranging gives us:
\[ 50 L_f - 30 C = 4000 \] (Equation 1)

### Step 4: Calculate heat lost when 80 g of ice is added
1. **Total mass of water after first melting**:
- Mass of water after melting 50 g of ice = 200 g + 50 g = 250 g

2. **Heat lost by the new total water**:
- New initial temperature = 40°C
- Final temperature = 10°C
- Change in temperature (ΔT_w2) = 40°C - 10°C = 30°C
- Heat lost by the new total water (Q_w2) = 250 g * 1 cal/g°C * 30°C = 7500 cal

3. **Heat lost by the flask**:
- Change in temperature (ΔT_f2) = 40°C - 10°C = 30°C
- Heat lost by the flask (Q_f2) = C * 30°C

4. **Heat gained by the additional ice**:
- Mass of additional ice (m_i2) = 80 g
- Heat gained by the additional ice (Q_i2) = m_i2 * L_f + m_i2 * s * ΔT_i2
- Change in temperature for additional ice (ΔT_i2) = 10°C - 0°C = 10°C
- Heat gained by the additional ice (Q_i2) = 80 g * L_f + 80 g * 1 cal/g°C * 10°C = 80 L_f + 800 cal

### Step 5: Set up the equation for the second scenario
From the conservation of energy:
\[ Q_w2 + Q_f2 = Q_i2 \]
Substituting the values:
\[ 7500 + C * 30 = 80 L_f + 800 \]
Rearranging gives us:
\[ 80 L_f - 30 C = 6700 \] (Equation 2)

### Step 6: Solve the two equations simultaneously
We have:
1. \( 50 L_f - 30 C = 4000 \) (Equation 1)
2. \( 80 L_f - 30 C = 6700 \) (Equation 2)

Subtract Equation 1 from Equation 2:
\[ (80 L_f - 30 C) - (50 L_f - 30 C) = 6700 - 4000 \]
This simplifies to:
\[ 30 L_f = 2700 \]
Thus,
\[ L_f = 90 \text{ cal/g} \]

### Final Answer:
The latent heat of fusion of ice is **90 cal/g**.

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