Steam at 373K is passed through a tube of radius 50 cm and length 3m. If thickness of the tube be 2 mm and conductivity of its material be `2xx10^(-4)`/ cal/s-cm-K, the rate of loss of heat in `Js^(-1)` is (The outside temperature is 282K).

Options

A. 36021J/s
B. 26026J/s
C. 46026J/s
D. 56026J/s

Answer

To solve the problem of calculating the rate of heat loss through a tube, we will use the formula for heat conduction through a cylindrical wall. Here’s a step-by-step breakdown:

### Step 1: Identify the given values
- **Inner temperature (T1)** = 373 K (temperature of steam)
- **Outer temperature (T2)** = 282 K (outside temperature)
- **Radius of the tube (r)** = 50 cm = 0.5 m
- **Length of the tube (L)** = 3 m
- **Thickness of the tube (d)** = 2 mm = 0.002 m
- **Thermal conductivity (k)** = \(2 \times 10^{-4}\) cal/s·cm·K = \(2 \times 10^{-4} \times 4184\) J/s·m·K (since 1 cal = 4184 J and 1 cm = 0.01 m)

### Step 2: Convert thermal conductivity to appropriate units
\[ k = 2 \times 10^{-4} \text{ cal/s·cm·K} = 2 \times 10^{-4} \times 4184 \text{ J/s·m·K} \]
Calculating this gives:
\[ k = 0.0008368 \text{ J/s·m·K} \]

### Step 3: Calculate the area through which heat is conducted
The area \(A\) of the cylindrical surface through which heat is conducted is given by:
\[ A = 2 \pi r L \]
Substituting the values:
\[ A = 2 \pi (0.5 \text{ m}) (3 \text{ m}) = 3 \pi \text{ m}^2 \]

### Step 4: Calculate the temperature difference
The temperature difference \(\Delta T\) is:
\[ \Delta T = T1 - T2 = 373 \text{ K} - 282 \text{ K} = 91 \text{ K} \]

### Step 5: Calculate the rate of heat loss using Fourier's law
The rate of heat transfer \(Q/t\) through the tube can be calculated using the formula:
\[ \frac{Q}{t} = \frac{k A \Delta T}{d} \]
Substituting the values:
\[ \frac{Q}{t} = \frac{(0.0008368 \text{ J/s·m·K}) (3 \pi \text{ m}^2) (91 \text{ K})}{0.002 \text{ m}} \]

### Step 6: Simplify and calculate
Calculating the above expression:
1. Calculate \(3 \pi\):
\[ 3 \pi \approx 9.4248 \text{ m}^2 \]
2. Now substitute:
\[ \frac{Q}{t} = \frac{(0.0008368) (9.4248) (91)}{0.002} \]
3. Calculate:
\[ \frac{Q}{t} \approx \frac{0.0008368 \times 9.4248 \times 91}{0.002} \]
\[ \approx \frac{0.0008368 \times 856.56}{0.002} \]
\[ \approx \frac{0.7164}{0.002} \approx 358.2 \text{ J/s} \]

### Final Answer
The rate of loss of heat is approximately **358.2 J/s**.

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