A platinum balls of mas `100g` is remoived from a furnace and immersed in a copper vessel of mass `100g` containing water of mass `390g` at `30^(@)C`. The temperature of water rises to `40^(@)C`. Caltulate the temperature of the furnace. (Given that the specific heat of platinum `=168kg^(-1)K^(-1)`, specific heat capacity of copper `=420Jkg^(-1)K^(-1)` and specific heat capacity of water `=4200Jkg^(-1)K^(-1))`

Answer

To solve the problem, we will use the principle of calorimetry, which states that the heat lost by the platinum ball will be equal to the heat gained by the water and the copper vessel.

### Step 1: Identify the known values
- Mass of platinum ball, \( m_{Pt} = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Specific heat capacity of platinum, \( c_{Pt} = 168 \, \text{J/kg/K} \)
- Mass of copper vessel, \( m_{Cu} = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Specific heat capacity of copper, \( c_{Cu} = 420 \, \text{J/kg/K} \)
- Mass of water, \( m_{water} = 390 \, \text{g} = 0.39 \, \text{kg} \)
- Specific heat capacity of water, \( c_{water} = 4200 \, \text{J/kg/K} \)
- Initial temperature of water, \( T_{initial} = 30^\circ C \)
- Final temperature of water, \( T_{final} = 40^\circ C \)

### Step 2: Calculate the heat gained by water and copper vessel
The heat gained by the water can be calculated using the formula:
\[ Q_{water} = m_{water} \cdot c_{water} \cdot (T_{final} - T_{initial}) \]
Substituting the values:
\[ Q_{water} = 0.39 \, \text{kg} \cdot 4200 \, \text{J/kg/K} \cdot (40 - 30) \]
\[ Q_{water} = 0.39 \cdot 4200 \cdot 10 = 16380 \, \text{J} \]

The heat gained by the copper vessel is:
\[ Q_{Cu} = m_{Cu} \cdot c_{Cu} \cdot (T_{final} - T_{initial}) \]
Substituting the values:
\[ Q_{Cu} = 0.1 \, \text{kg} \cdot 420 \, \text{J/kg/K} \cdot (40 - 30) \]
\[ Q_{Cu} = 0.1 \cdot 420 \cdot 10 = 420 \, \text{J} \]

### Step 3: Calculate total heat gained
Total heat gained by water and copper vessel:
\[ Q_{total} = Q_{water} + Q_{Cu} = 16380 \, \text{J} + 420 \, \text{J} = 16800 \, \text{J} \]

### Step 4: Set up the equation for heat lost by platinum
The heat lost by the platinum ball is given by:
\[ Q_{Pt} = m_{Pt} \cdot c_{Pt} \cdot (T - T_{final}) \]
Setting the heat lost equal to the heat gained:
\[ m_{Pt} \cdot c_{Pt} \cdot (T - 40) = Q_{total} \]
Substituting the known values:
\[ 0.1 \cdot 168 \cdot (T - 40) = 16800 \]

### Step 5: Solve for T
\[ 16.8 \cdot (T - 40) = 16800 \]
\[ T - 40 = \frac{16800}{16.8} \]
\[ T - 40 = 1000 \]
\[ T = 1000 + 40 = 1040^\circ C \]

### Final Answer
The temperature of the furnace is \( T = 1040^\circ C \).

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