An aluminium vessel of mass `0.5kg` contains `0.2kg` of water at `20^(@)C` A block of iron of mass `0.2kg at 100^(@)C` is gently put into the water .Find the equilibrium temperature of the mixture,Specific beat capactities of aluminium , iron and water are `910 J kg^(-1)K^(-1) 470J kg^(-1)K^(-1) and 420J kg^(-1)K^(-1)` respectively

Explanation

To find the equilibrium temperature of the mixture containing an aluminum vessel, water, and an iron block, we can follow these steps:

### Step 1: Identify the given data
- Mass of aluminum vessel, \( m_{Al} = 0.5 \, \text{kg} \)
- Mass of water, \( m_{water} = 0.2 \, \text{kg} \)
- Mass of iron block, \( m_{Fe} = 0.2 \, \text{kg} \)
- Initial temperature of water and aluminum, \( T_{initial} = 20^\circ C = 293 \, \text{K} \)
- Initial temperature of iron, \( T_{Fe} = 100^\circ C = 373 \, \text{K} \)
- Specific heat capacity of aluminum, \( c_{Al} = 910 \, \text{J/kg K} \)
- Specific heat capacity of iron, \( c_{Fe} = 470 \, \text{J/kg K} \)
- Specific heat capacity of water, \( c_{water} = 4200 \, \text{J/kg K} \)

### Step 2: Set up the heat transfer equations
The heat lost by the iron block will be equal to the heat gained by the water and the aluminum vessel at equilibrium.

1. **Heat lost by iron**:
\[ Q_{Fe} = m_{Fe} \cdot c_{Fe} \cdot (T_{Fe} - T) \]
\[ Q_{Fe} = 0.2 \cdot 470 \cdot (373 - T) \]

2. **Heat gained by water**:
\[ Q_{water} = m_{water} \cdot c_{water} \cdot (T - T_{initial}) \]
\[ Q_{water} = 0.2 \cdot 4200 \cdot (T - 293) \]

3. **Heat gained by aluminum**:
\[ Q_{Al} = m_{Al} \cdot c_{Al} \cdot (T - T_{initial}) \]
\[ Q_{Al} = 0.5 \cdot 910 \cdot (T - 293) \]

### Step 3: Combine the heat equations
Since the heat lost by iron equals the total heat gained by water and aluminum:
\[ Q_{Fe} = Q_{water} + Q_{Al} \]
Substituting the expressions we derived:
\[ 0.2 \cdot 470 \cdot (373 - T) = 0.2 \cdot 4200 \cdot (T - 293) + 0.5 \cdot 910 \cdot (T - 293) \]

### Step 4: Simplify and solve for \( T \)
1. Calculate the left side:
\[ 0.2 \cdot 470 \cdot (373 - T) = 94 \cdot (373 - T) \]

2. Calculate the right side:
\[ 0.2 \cdot 4200 \cdot (T - 293) = 840 \cdot (T - 293) \]
\[ 0.5 \cdot 910 \cdot (T - 293) = 455 \cdot (T - 293) \]
Combine these:
\[ Q_{total} = 840 \cdot (T - 293) + 455 \cdot (T - 293) = (840 + 455) \cdot (T - 293) = 1295 \cdot (T - 293) \]

3. Set the equation:
\[ 94 \cdot (373 - T) = 1295 \cdot (T - 293) \]

4. Expand and rearrange:
\[ 35042 - 94T = 1295T - 379735 \]
\[ 35042 + 379735 = 1295T + 94T \]
\[ 414777 = 1389T \]
\[ T = \frac{414777}{1389} \approx 299.33 \, \text{K} \]

### Step 5: Convert to Celsius
\[ T_{final} = 299.33 - 273 \approx 26.33^\circ C \]

### Final Answer
The equilibrium temperature of the mixture is approximately \( 26.33^\circ C \).