`50g` of ice at `0^(@)C` is mixed with `200g` of water at `0^(@)C.6` kcal heat is given to system [Ice +water]. Find the temperature (in `.^(@)C`) of the system.

Explanation

To solve the problem step by step, we will follow the principles of calorimetry, considering the heat exchange between the ice and water.

### Step 1: Understand the System
We have:
- 50 g of ice at 0°C
- 200 g of water at 0°C
- 6 kcal of heat is added to the system.

### Step 2: Define Variables
Let:
- \( T \) = final temperature of the system (in °C)
- \( x \) = heat absorbed by the ice (in kcal)
- Heat absorbed by the water = \( 6 - x \) kcal

### Step 3: Calculate Heat Absorbed by Ice
The heat absorbed by the ice consists of two parts:
1. Melting the ice to water at 0°C.
2. Heating the resulting water from 0°C to \( T \).

The heat required to melt the ice:
\[ Q_{\text{ice}} = m \cdot L = 50 \, \text{g} \cdot 80 \, \text{cal/g} = 4000 \, \text{cal} = 4 \, \text{kcal} \]

The heat required to raise the temperature of the melted ice (now water) from 0°C to \( T \):
\[ Q_{\text{water}} = m \cdot c \cdot \Delta T = 50 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (T - 0) = 50T \, \text{cal} = 0.05T \, \text{kcal} \]

Thus, the total heat absorbed by the ice is:
\[ x = 4 + 0.05T \]

### Step 4: Calculate Heat Absorbed by Water
The heat absorbed by the water as it warms from 0°C to \( T \):
\[ Q_{\text{water}} = m \cdot c \cdot \Delta T = 200 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (T - 0) = 200T \, \text{cal} = 0.2T \, \text{kcal} \]

Thus, the heat absorbed by the water is:
\[ 6 - x = 0.2T \]

### Step 5: Set Up the Equation
Now we have two equations for \( x \):
1. \( x = 4 + 0.05T \)
2. \( 6 - x = 0.2T \)

Substituting the first equation into the second:
\[ 6 - (4 + 0.05T) = 0.2T \]
\[ 6 - 4 - 0.05T = 0.2T \]
\[ 2 = 0.2T + 0.05T \]
\[ 2 = 0.25T \]

### Step 6: Solve for \( T \)
\[ T = \frac{2}{0.25} = 8 \, \text{°C} \]

### Conclusion
The final temperature of the system is \( T = 8 \, \text{°C} \).

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