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 heat capacities of aluminium , iron and water are `910 J kg^(-1)K^(-1) 470J kg^(-1)K^(-1) and 4200J kg^(-1)K^(-1)` respectively

Explanation

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

### Step 1: Identify the known quantities
- 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 aluminum and water, \( T_{initial} = 20^\circ C = 293 \, \text{K} \)
- Initial temperature of iron, \( T_{Fe} = 100^\circ C = 373 \, \text{K} \)
- Specific heat capacity of aluminum, \( c_{Al} = 910 \, \text{J kg}^{-1} \text{K}^{-1} \)
- Specific heat capacity of water, \( c_{water} = 4200 \, \text{J kg}^{-1} \text{K}^{-1} \)
- Specific heat capacity of iron, \( c_{Fe} = 470 \, \text{J kg}^{-1} \text{K}^{-1} \)

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

Using the formula for heat transfer:
\[ Q = mc\Delta T \]

The heat lost by the iron block is:
\[ Q_{Fe} = m_{Fe} \cdot c_{Fe} \cdot (T_{Fe} - T) \]
Substituting the values:
\[ Q_{Fe} = 0.2 \cdot 470 \cdot (373 - T) \]

The heat gained by the water is:
\[ Q_{water} = m_{water} \cdot c_{water} \cdot (T - T_{initial}) \]
Substituting the values:
\[ Q_{water} = 0.2 \cdot 4200 \cdot (T - 293) \]

The heat gained by the aluminum vessel is:
\[ Q_{Al} = m_{Al} \cdot c_{Al} \cdot (T - T_{initial}) \]
Substituting the values:
\[ Q_{Al} = 0.5 \cdot 910 \cdot (T - 293) \]

### Step 3: Combine the equations
Since the heat lost by the iron is equal to the total heat gained by the 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 the equation
Expanding both sides:
\[ 94 \cdot (373 - T) = 840 \cdot (T - 293) + 455 \cdot (T - 293) \]

Calculating the right side:
\[ 94 \cdot (373 - T) = (840 + 455) \cdot (T - 293) \]
\[ 94 \cdot (373 - T) = 1295 \cdot (T - 293) \]

### Step 5: Solve for T
Expanding both sides:
\[ 35082 - 94T = 1295T - 379735 \]

Bringing all terms involving T to one side:
\[ 35082 + 379735 = 1295T + 94T \]
\[ 414817 = 1389T \]

Now, solving for T:
\[ T = \frac{414817}{1389} \approx 299.0 \, \text{K} \]

### Step 6: Convert to Celsius
To convert from Kelvin to Celsius:
\[ T_{equilibrium} = 299.0 \, \text{K} - 273.15 \approx 25.85^\circ C \]

### Final Answer
The equilibrium temperature of the mixture is approximately \( 25.85^\circ C \).

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