Steam at `120^(@)C` is continuously passed through a `50-cm` long rubber tube of inner and outer radii `1.0cm` and `1.2cm` . The room temperature is `30^(@)C` . Calculate the rate of heat flow through the walls of the tube. Thermal conductivity of rubber `= 0.15Js^(-1)m^(-1)C^(-1)` .

Explanation

To solve the problem of calculating the rate of heat flow through the walls of a rubber tube, we will use the formula for heat conduction through a cylindrical surface. Here’s the step-by-step solution:

### Step 1: Identify the given values
- Temperature of steam, \( T_1 = 120^\circ C \)
- Room temperature, \( T_2 = 30^\circ C \)
- Length of the tube, \( l = 50 \, \text{cm} = 0.50 \, \text{m} \)
- Inner radius of the tube, \( r_1 = 1.0 \, \text{cm} = 0.01 \, \text{m} \)
- Outer radius of the tube, \( r_2 = 1.2 \, \text{cm} = 0.012 \, \text{m} \)
- Thermal conductivity of rubber, \( k = 0.15 \, \text{W/m} \cdot \text{°C} \)

### Step 2: Calculate the temperature difference
The temperature difference \( \Delta T \) between the steam and the room temperature is calculated as:
\[ \Delta T = T_1 - T_2 = 120^\circ C - 30^\circ C = 90^\circ C \]

### Step 3: Use the formula for heat flow through a cylindrical wall
The formula for the rate of heat flow \( Q \) through the cylindrical wall is given by:
\[ Q = \frac{2 \pi k l (T_1 - T_2)}{\ln \left( \frac{r_2}{r_1} \right)} \]

### Step 4: Substitute the values into the formula
Substituting the known values into the formula:
\[ Q = \frac{2 \pi (0.15) (0.50) (90)}{\ln \left( \frac{0.012}{0.01} \right)} \]

### Step 5: Calculate the logarithm
Calculate the logarithm:
\[ \ln \left( \frac{0.012}{0.01} \right) = \ln (1.2) \approx 0.1823 \]

### Step 6: Calculate the rate of heat flow
Now substitute this back into the equation:
\[ Q = \frac{2 \pi (0.15) (0.50) (90)}{0.1823} \]
Calculating the numerator:
\[ 2 \pi (0.15) (0.50) (90) \approx 42.411 \, \text{W} \]
Now divide by the logarithm:
\[ Q \approx \frac{42.411}{0.1823} \approx 232.7 \, \text{W} \]

### Step 7: Final answer
The rate of heat flow through the walls of the tube is approximately:
\[ Q \approx 233 \, \text{W} \]