Find the final temeprature and composition of the mixture of 1 kg of ice at `-10^(@)C` and 4.4 kg of water at `30^(@)C`. Given that specific heat of water is `4200J//kg^(@)C` and that of ice is `2100J//kg^(@)C` and latent heat of fusion of ice is `3.36xx10^(5)`J/kg

Answer

To solve the problem of finding the final temperature and composition of the mixture of 1 kg of ice at -10°C and 4.4 kg of water at 30°C, we will follow these steps:

### Step 1: Calculate the energy required to raise the temperature of ice from -10°C to 0°C
The formula to calculate the heat energy (q) required to change the temperature is:
\[ q = m \cdot s \cdot \Delta T \]
Where:
- \( m \) = mass of ice = 1 kg
- \( s \) = specific heat of ice = 2100 J/kg°C
- \( \Delta T \) = change in temperature = \( 0 - (-10) = 10°C \)

Substituting the values:
\[ q_{\text{ice}} = 1 \, \text{kg} \cdot 2100 \, \text{J/kg°C} \cdot 10°C = 21000 \, \text{J} \]

### Step 2: Calculate the energy released by water when it cools from 30°C to 0°C
Using the same formula:
- \( m \) = mass of water = 4.4 kg
- \( s \) = specific heat of water = 4200 J/kg°C
- \( \Delta T \) = change in temperature = \( 30 - 0 = 30°C \)

Substituting the values:
\[ q_{\text{water}} = 4.4 \, \text{kg} \cdot 4200 \, \text{J/kg°C} \cdot 30°C = 554400 \, \text{J} \]

### Step 3: Determine if the energy from water is sufficient to convert ice to water
The energy required to convert 1 kg of ice at 0°C to water at 0°C is given by:
\[ q_{\text{fusion}} = m \cdot L_f \]
Where:
- \( L_f \) = latent heat of fusion of ice = \( 3.36 \times 10^5 \, \text{J/kg} \)

Substituting the values:
\[ q_{\text{fusion}} = 1 \, \text{kg} \cdot 3.36 \times 10^5 \, \text{J/kg} = 336000 \, \text{J} \]

### Step 4: Calculate the total energy available from water
The total energy available from the water when it cools down to 0°C is:
\[ q_{\text{total}} = q_{\text{water}} - q_{\text{ice}} - q_{\text{fusion}} \]
Substituting the values:
\[ q_{\text{total}} = 554400 \, \text{J} - 21000 \, \text{J} - 336000 \, \text{J} \]
\[ q_{\text{total}} = 554400 \, \text{J} - 357000 \, \text{J} = 197400 \, \text{J} \]

### Step 5: Calculate the final temperature of the mixture
Now, we will use the remaining energy to raise the temperature of the resulting water mixture. The mass of the mixture is:
\[ m_{\text{mixture}} = 4.4 \, \text{kg} + 1 \, \text{kg} = 5.4 \, \text{kg} \]

Using the formula for heat again:
\[ q_{\text{remaining}} = m_{\text{mixture}} \cdot s \cdot \Delta T \]
Where:
- \( s \) = specific heat of water = 4200 J/kg°C
- \( \Delta T \) = final temperature - initial temperature (0°C)

Rearranging for \( \Delta T \):
\[ \Delta T = \frac{q_{\text{remaining}}}{m_{\text{mixture}} \cdot s} \]
Substituting the values:
\[ \Delta T = \frac{197400 \, \text{J}}{5.4 \, \text{kg} \cdot 4200 \, \text{J/kg°C}} \]
\[ \Delta T = \frac{197400}{22680} \approx 8.7°C \]

### Step 6: Calculate the final temperature
The final temperature \( T_f \) is:
\[ T_f = 0°C + 8.7°C = 8.7°C \]

### Conclusion
The final temperature of the mixture is **8.7°C** and the composition is **5.4 kg of water**.

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