A boiler is made of a copper plate `2.4mm` thick with an inside coating of a `0.2mm` thick layer of tin The surface area exposed to gases at `700^(@)C` is `400cm^(2)` The maximum amount of steam that could be generated per hour at atmospheric pressure is
`({:(K_(cu)=0.9cal//cm-s-^(0)&k_("tin"=0.15cal//cm/s/^(0)C),,,,),(andL_(steam)=540cal//g,,,,):})` .

Options

A. 5000 kg
B. 1000 kg
C. 4000 kg
D. 200 kg

Answer

To solve the problem of determining the maximum amount of steam that could be generated per hour by the boiler, we will follow these steps:

### Step 1: Identify the given data
- Thickness of copper plate, \( L_{cu} = 2.4 \, \text{mm} = 0.24 \, \text{cm} \)
- Thickness of tin coating, \( L_{tin} = 0.2 \, \text{mm} = 0.02 \, \text{cm} \)
- Thermal conductivity of copper, \( K_{cu} = 0.9 \, \text{cal/cm/s/°C} \)
- Thermal conductivity of tin, \( K_{tin} = 0.15 \, \text{cal/cm/s/°C} \)
- Surface area exposed to gases, \( A = 400 \, \text{cm}^2 \)
- Temperature of gases, \( T_{g} = 700 \, \text{°C} \)
- Temperature of water inside the boiler, \( T_{w} = 100 \, \text{°C} \)
- Latent heat of steam, \( L_{steam} = 540 \, \text{cal/g} \)

### Step 2: Calculate the total thermal resistance
The total thermal resistance \( R \) of the system can be calculated using the formula:
\[ R = R_{cu} + R_{tin} \]
Where:
\[ R_{cu} = \frac{L_{cu}}{K_{cu} \cdot A} \quad \text{and} \quad R_{tin} = \frac{L_{tin}}{K_{tin} \cdot A} \]

Substituting the values:
\[ R_{cu} = \frac{0.24 \, \text{cm}}{0.9 \, \text{cal/cm/s/°C} \cdot 400 \, \text{cm}^2} = \frac{0.24}{360} = 0.0006667 \, \text{s/°C} \]
\[ R_{tin} = \frac{0.02 \, \text{cm}}{0.15 \, \text{cal/cm/s/°C} \cdot 400 \, \text{cm}^2} = \frac{0.02}{60} = 0.0003333 \, \text{s/°C} \]

Now, adding them together:
\[ R = R_{cu} + R_{tin} = 0.0006667 + 0.0003333 = 0.001 \, \text{s/°C} \]

### Step 3: Calculate the heat transfer rate (Q)
Using the formula for heat transfer:
\[ Q = \frac{T_{g} - T_{w}}{R} \]
Substituting the values:
\[ Q = \frac{700 - 100}{0.001} = \frac{600}{0.001} = 600,000 \, \text{cal/s} \]

### Step 4: Calculate the mass of steam generated per second
To find the mass of steam generated per second, we use:
\[ \text{mass per second} = \frac{Q}{L_{steam}} \]
Substituting the values:
\[ \text{mass per second} = \frac{600,000 \, \text{cal/s}}{540 \, \text{cal/g}} \approx 1111.11 \, \text{g/s} \]

### Step 5: Convert to mass per hour
To find the mass of steam generated in one hour:
\[ \text{mass per hour} = 1111.11 \, \text{g/s} \times 3600 \, \text{s} = 4,000,000 \, \text{g} = 4000 \, \text{kg} \]

### Conclusion
The maximum amount of steam that could be generated per hour at atmospheric pressure is approximately **4000 kg**.

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