An iron boiler with walls 1.25 cm thick contains water at atmospheric pressure. The heated surface is `2.5 m^(2)` in area and the temperature of the underside is `120^(@)C.` Thermal conductivity of iron is `20 cal s^(-1) m^(-1) K^(-1)` and the latent heat of evaporation of water `536 xx10^(3) cal kg^(-1) `. Find the mass of water evaporated per hour.

Answer

To solve the problem of finding the mass of water evaporated per hour from the iron boiler, we will follow these steps:

### Step 1: Identify the Given Data
- Thickness of the boiler wall, \( d = 1.25 \, \text{cm} = 0.0125 \, \text{m} \)
- Area of the heated surface, \( A = 2.5 \, \text{m}^2 \)
- Temperature of the underside, \( T_1 = 120^\circ C \)
- Temperature of the water, \( T_2 = 100^\circ C \)
- Thermal conductivity of iron, \( K = 20 \, \text{cal} \, \text{s}^{-1} \, \text{m}^{-1} \, \text{K}^{-1} \)
- Latent heat of evaporation of water, \( L = 536 \times 10^3 \, \text{cal/kg} \)

### Step 2: Calculate the Temperature Difference
The temperature difference between the underside of the boiler and the water is:
\[ \Delta T = T_1 - T_2 = 120^\circ C - 100^\circ C = 20 \, \text{K} \]

### Step 3: Calculate the Rate of Heat Transfer
Using the formula for heat conduction:
\[ \frac{dQ}{dt} = \frac{K \cdot A \cdot \Delta T}{d} \]
Substituting the values:
\[ \frac{dQ}{dt} = \frac{20 \, \text{cal/s/m/K} \cdot 2.5 \, \text{m}^2 \cdot 20 \, \text{K}}{0.0125 \, \text{m}} \]

Calculating this:
\[ \frac{dQ}{dt} = \frac{1000 \, \text{cal/s}}{0.0125} = 80000 \, \text{cal/s} \]

### Step 4: Convert to Heat Transfer per Hour
To find the total heat transferred in one hour:
\[ Q = \frac{dQ}{dt} \times 3600 \, \text{s} = 80000 \, \text{cal/s} \times 3600 \, \text{s} = 288000000 \, \text{cal} \]

### Step 5: Calculate the Mass of Water Evaporated
Using the relationship between heat and mass for evaporation:
\[ Q = mL \implies m = \frac{Q}{L} \]
Substituting the values:
\[ m = \frac{288000000 \, \text{cal}}{536 \times 10^3 \, \text{cal/kg}} = \frac{288000000}{536000} \approx 537.31 \, \text{kg} \]

### Final Answer
The mass of water evaporated per hour is approximately **537.31 kg**.

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