An iron boiler is 1 cm thick and has a heating area `2.5 m^(2)`. The two surface of the boiler are at `230^(@)C` and `100^(@)C` respectively. If the latent heat of the steam is `540 kcal kg^(-1)` and thermal conductivity of iron is `1.6xx10^(-2)K cal s^(-1) m^(-1)K^(-1)`, then how much water will be evaporated into steam per minute ?

Explanation

To solve the problem, we need to calculate the amount of water that will evaporate into steam per minute using the given data. The steps are as follows:

### Step 1: Identify the given data
- Thickness of the boiler, \( \Delta x = 1 \, \text{cm} = 0.01 \, \text{m} \)
- Heating area, \( A = 2.5 \, \text{m}^2 \)
- Temperature difference, \( \Delta T = 230^\circ C - 100^\circ C = 130 \, \text{K} \)
- Latent heat of steam, \( L = 540 \, \text{kcal/kg} \)
- Thermal conductivity of iron, \( k = 1.6 \times 10^{-2} \, \text{kcal/s m K} \)

### Step 2: Calculate the rate of heat transfer (\( \frac{\Delta Q}{\Delta t} \))
Using the formula for heat conduction:
\[ \frac{\Delta Q}{\Delta t} = \frac{k \cdot A \cdot \Delta T}{\Delta x} \]
Substituting the values:
\[ \frac{\Delta Q}{\Delta t} = \frac{(1.6 \times 10^{-2} \, \text{kcal/s m K}) \cdot (2.5 \, \text{m}^2) \cdot (130 \, \text{K})}{0.01 \, \text{m}} \]

### Step 3: Perform the calculation
Calculating the numerator:
\[ 1.6 \times 10^{-2} \cdot 2.5 \cdot 130 = 5.2 \, \text{kcal/s} \]
Now, divide by the thickness:
\[ \frac{5.2}{0.01} = 520 \, \text{kcal/s} \]

### Step 4: Calculate the mass of water evaporated per minute
The heat required to evaporate water can be expressed as:
\[ \Delta Q = m \cdot L \]
Where \( m \) is the mass of water evaporated. Rearranging gives:
\[ m = \frac{\Delta Q}{L} \]
Now, we need to find the total heat transferred in one minute:
\[ \Delta Q = \frac{\Delta Q}{\Delta t} \times 60 \, \text{s} = 520 \, \text{kcal/s} \times 60 \, \text{s} = 31200 \, \text{kcal} \]

Now substituting into the mass equation:
\[ m = \frac{31200 \, \text{kcal}}{540 \, \text{kcal/kg}} = 57.78 \, \text{kg} \]

### Conclusion
The amount of water that will evaporate into steam per minute is approximately **57.78 kg**.

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