Water is boiled in a container having a bottom of surface area `25cm^(2)` , thickness `1.0mm` and thermal conductivity `50Wm^(-1)C^(-1)` , 100g of water is converted into steam per minute in the steady state after the boiling starts. Assuming that no heat is lost to the atmosphere, calculate the temperature of the lower surface of the bottom. Latent heat of vaporization of water `=2.26xx10^(6)Jkg^(-1)` .

Explanation

To solve the problem step by step, we will follow the principles of heat transfer and the properties of materials involved.

### Given Data:
- Surface area of the container, \( A = 25 \, \text{cm}^2 = 25 \times 10^{-4} \, \text{m}^2 = 2.5 \times 10^{-2} \, \text{m}^2 \)
- Thickness of the container, \( d = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} \)
- Thermal conductivity, \( K = 50 \, \text{W/m°C} \)
- Mass of water converted to steam per minute, \( m = 100 \, \text{g/min} = \frac{100 \times 10^{-3}}{60} \, \text{kg/s} = \frac{1}{600} \, \text{kg/s} \)
- Latent heat of vaporization, \( L = 2.26 \times 10^6 \, \text{J/kg} \)

### Step 1: Calculate the rate of heat required to convert water to steam.
The heat required per second (\( \dot{Q} \)) to convert the mass of water to steam can be calculated using the formula:
\[ \dot{Q} = \dot{m} \times L \]
Where \( \dot{m} \) is the mass flow rate in kg/s.

Substituting the values:
\[ \dot{Q} = \left(\frac{1}{600}\right) \times (2.26 \times 10^6) = \frac{2.26 \times 10^6}{600} \approx 3766.67 \, \text{W} \]

### Step 2: Use the heat conduction formula to find the temperature of the lower surface.
The rate of heat conduction through the container can be expressed as:
\[ \dot{Q} = \frac{K \cdot A \cdot (T_s - T_w)}{d} \]
Where:
- \( T_s \) is the temperature of the lower surface of the container.
- \( T_w \) is the temperature of the water (which is \( 100 \, °C \)).

Rearranging the formula to find \( T_s \):
\[ T_s = T_w + \frac{\dot{Q} \cdot d}{K \cdot A} \]

### Step 3: Substitute the known values into the equation.
Substituting the values we have:
\[ T_s = 100 + \frac{3766.67 \cdot (1.0 \times 10^{-3})}{50 \cdot (2.5 \times 10^{-2})} \]

Calculating the denominator:
\[ 50 \cdot (2.5 \times 10^{-2}) = 1.25 \, \text{W/°C} \]

Now substituting back:
\[ T_s = 100 + \frac{3766.67 \cdot (1.0 \times 10^{-3})}{1.25} \]
\[ T_s = 100 + \frac{3.76667}{1.25} \approx 100 + 3.01334 \approx 103.01 \, °C \]

### Step 4: Final Calculation
Calculating the final temperature:
\[ T_s \approx 103.01 \, °C \]

### Conclusion
The temperature of the lower surface of the bottom of the container is approximately \( 103.01 \, °C \).