About 5 g water at `30^(@)C` and `5 g` is at `-20^(@)C` are mixed in a container of negligible heat capacity. Find the final temperature and composition of the mixture. Given that specific heat of water is `4200J//kg^(@)C` and that of ice of `2100//kg^(@)C` and latent heat of ice is `3.36xx10^(5)`J/Kg

Answer

To find the final temperature and composition of the mixture when 5 g of water at \(30^\circ C\) is mixed with 5 g of ice at \(-20^\circ C\), we can follow these steps:

### Step 1: Convert the specific heats and latent heat to appropriate units
Given:
- Specific heat of water, \(S_w = 4200 \, \text{J/kg} \cdot \text{C} = 4.2 \, \text{J/g} \cdot \text{C} = 1 \, \text{cal/g} \cdot \text{C}\)
- Specific heat of ice, \(S_i = 2100 \, \text{J/kg} \cdot \text{C} = 2.1 \, \text{J/g} \cdot \text{C} = 0.5 \, \text{cal/g} \cdot \text{C}\)
- Latent heat of fusion of ice, \(L_f = 3.36 \times 10^5 \, \text{J/kg} = 80 \, \text{cal/g}\)

### Step 2: Calculate the heat released by the water when it cools to \(0^\circ C\)
The water will lose heat as it cools from \(30^\circ C\) to \(0^\circ C\):
\[ Q_{\text{water}} = m \cdot S_w \cdot \Delta T = 5 \, \text{g} \cdot 1 \, \text{cal/g} \cdot \text{C} \cdot (30 - 0) \, \text{C} = 5 \cdot 30 = 150 \, \text{cal} \]

### Step 3: Calculate the heat required to warm the ice to \(0^\circ C\)
The ice will gain heat as it warms from \(-20^\circ C\) to \(0^\circ C\):
\[ Q_{\text{ice}} = m \cdot S_i \cdot \Delta T = 5 \, \text{g} \cdot 0.5 \, \text{cal/g} \cdot \text{C} \cdot (0 - (-20)) \, \text{C} = 5 \cdot 0.5 \cdot 20 = 50 \, \text{cal} \]

### Step 4: Determine the remaining heat after warming the ice
The water releases \(150 \, \text{cal}\) and the ice requires \(50 \, \text{cal}\) to reach \(0^\circ C\):
\[ \text{Remaining heat} = Q_{\text{water}} - Q_{\text{ice}} = 150 \, \text{cal} - 50 \, \text{cal} = 100 \, \text{cal} \]

### Step 5: Calculate how much ice can be converted to water with the remaining heat
The remaining heat will be used to convert some of the ice at \(0^\circ C\) into water:
\[ Q = m \cdot L_f \implies m = \frac{Q}{L_f} = \frac{100 \, \text{cal}}{80 \, \text{cal/g}} = 1.25 \, \text{g} \]

### Step 6: Determine the final composition of the mixture
- Initial mass of ice = \(5 \, \text{g}\)
- Mass of ice converted to water = \(1.25 \, \text{g}\)
- Remaining mass of ice = \(5 \, \text{g} - 1.25 \, \text{g} = 3.75 \, \text{g}\)
- Initial mass of water = \(5 \, \text{g}\)
- Final mass of water = \(5 \, \text{g} + 1.25 \, \text{g} = 6.25 \, \text{g}\)

### Final Result
The final temperature of the mixture is \(0^\circ C\), with:
- Mass of ice = \(3.75 \, \text{g}\)
- Mass of water = \(6.25 \, \text{g}\)