Water is being boiled in a flat bottomed kettle placed on a stove The area of the bottom is `300cm^(2)` and the thickness is `2mm` If the amount pf steam produced is `1gm min^(-1)` then the difference of the temperature between the inner and the outer surface of the bottom is (thermal conductivity of the matrial of the kettle `0.5calcm^(-1)C^(-1)` latent heat of the steam is equal to `540calg^(-1))` .

Options

A. `12^(@)C`
B. `1.2^(@)C`
C. `0.2^(@)C`
D. `0.012^(@)C`

Explanation

To solve the problem step by step, we need to find the temperature difference (ΔT) between the inner and outer surfaces of the kettle's bottom. Here’s how we can do it:

### Step 1: Understand the given data
- Area of the bottom of the kettle, A = 300 cm²
- Thickness of the kettle, B = 2 mm = 0.2 cm
- Rate of steam production, dm/dt = 1 g/min = 1/60 g/s
- Thermal conductivity of the kettle material, K = 0.5 cal/(cm·°C)
- Latent heat of steam, L = 540 cal/g

### Step 2: Calculate the heat required to produce steam
The heat required to produce steam at the rate of 1 g/min can be calculated using the formula:
\[ \frac{dq}{dt} = \frac{dm}{dt} \times L \]

Substituting the values:
\[ \frac{dq}{dt} = \left(\frac{1}{60} \text{ g/s}\right) \times 540 \text{ cal/g} \]
\[ \frac{dq}{dt} = \frac{540}{60} \text{ cal/s} = 9 \text{ cal/s} \]

### Step 3: Apply the formula for heat conduction
According to Fourier's law of heat conduction:
\[ \frac{dq}{dt} = K \cdot A \cdot \frac{\Delta T}{B} \]

Where:
- \( \Delta T \) is the temperature difference we need to find.

### Step 4: Rearranging the formula to find ΔT
Rearranging the equation gives:
\[ \Delta T = \frac{dq/dt \cdot B}{K \cdot A} \]

### Step 5: Substitute the known values
Substituting the known values into the equation:
- \( K = 0.5 \text{ cal/(cm·°C)} \)
- \( A = 300 \text{ cm}^2 \)
- \( B = 0.2 \text{ cm} \)
- \( \frac{dq}{dt} = 9 \text{ cal/s} \)

So,
\[ \Delta T = \frac{9 \text{ cal/s} \cdot 0.2 \text{ cm}}{0.5 \text{ cal/(cm·°C)} \cdot 300 \text{ cm}^2} \]

### Step 6: Calculate ΔT
Calculating the values:
\[ \Delta T = \frac{1.8 \text{ cal/(s·cm)}}{150 \text{ cal/(s·cm)}} \]
\[ \Delta T = \frac{1.8}{150} \]
\[ \Delta T = 0.012 \text{ °C} \]

### Conclusion
The temperature difference between the inner and outer surface of the kettle's bottom is **0.012 °C**.

---