Water is boiled in a rectangular steel tank of thickness 2 cm by a constant temperature furnace. Due to vaporisation, water level falls at a steady rate of 1 cm in 9 minutes. Calculate the temperature of the furnace. Given K for steel `0.2 cal s^(-1) m^(-1) .^(@)C^(-1)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We have a rectangular steel tank with:
- Thickness of the tank, \( x = 2 \) cm = 0.02 m
- Rate of water level falling = 1 cm in 9 minutes = \( \frac{1 \, \text{cm}}{9 \, \text{minutes}} = \frac{1 \, \text{cm}}{540 \, \text{seconds}} \)
- The thermal conductivity of steel, \( K = 0.2 \, \text{cal s}^{-1} \text{m}^{-1} \text{°C}^{-1} \)

### Step 2: Calculate the Volume of Water Vaporized
The volume of water that vaporizes in 9 minutes can be expressed as:
- Volume = Area \( A \) (in cm²) × height of water vaporized (in cm)
- Height of water vaporized in 9 minutes = 1 cm
- Therefore, Volume = \( A \times 1 \, \text{cm}^3 = A \, \text{cm}^3 \)

### Step 3: Convert Volume to Mass
Using the density of water (1 g/cm³):
- Mass of water vaporized = Volume × Density = \( A \, \text{g} \)

### Step 4: Calculate the Heat Required for Vaporization
The heat \( Q \) required to vaporize this mass of water can be calculated using the latent heat of vaporization \( L \):
- \( Q = mL = A \times L \)
- Assuming \( L \) (latent heat of vaporization) = 540 cal/g (for water), we get:
- \( Q = A \times 540 \, \text{cal} \)

### Step 5: Apply Fourier's Law of Heat Conduction
According to Fourier's law:
\[ Q = K \cdot A \cdot \frac{T_1 - T_2}{x} \cdot t \]
Where:
- \( T_1 \) = Temperature of the furnace
- \( T_2 \) = Temperature of the water (assumed to be 100°C)
- \( x \) = Thickness of the tank (0.02 m)
- \( t \) = Time (540 seconds)

### Step 6: Rearranging the Equation
Substituting \( Q \) into the equation:
\[ A \cdot 540 = K \cdot A \cdot \frac{T_1 - T_2}{x} \cdot t \]
Cancelling \( A \) from both sides (assuming \( A \neq 0 \)):
\[ 540 = K \cdot \frac{T_1 - T_2}{x} \cdot t \]

### Step 7: Substitute Known Values
Substituting \( K = 0.2 \, \text{cal s}^{-1} \text{m}^{-1} \text{°C}^{-1} \), \( x = 0.02 \, \text{m} \), and \( t = 540 \, \text{s} \):
\[ 540 = 0.2 \cdot \frac{T_1 - 100}{0.02} \cdot 540 \]

### Step 8: Simplify the Equation
Cancelling \( 540 \) from both sides:
\[ 1 = 0.2 \cdot \frac{T_1 - 100}{0.02} \]
\[ 1 = 10 (T_1 - 100) \]
\[ T_1 - 100 = 0.1 \]
\[ T_1 = 100 + 10 = 110 \, \text{°C} \]

### Final Answer
The temperature of the furnace is \( T_1 = 110 \, \text{°C} \).

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