Steam at 373 K is passed through a tube of radius 0.5 m and length 3m. If the thickness of the tube is 2 mm and `K= 2 xx 10^(-4)` in C.G.S. units find the rate of loss of heat in `Js^(-1)`. Room temperature is `9^(@)C`.

Explanation

To solve the problem of finding the rate of loss of heat through the tube, we will use the formula for heat conduction:

\[ \frac{Q}{T} = \frac{k \cdot A \cdot (T_1 - T_2)}{L} \]

Where:
- \( Q/T \) is the rate of heat flow (in Joules per second, or Watts),
- \( k \) is the thermal conductivity of the material (in CGS units),
- \( A \) is the cross-sectional area of the tube,
- \( T_1 \) and \( T_2 \) are the temperatures of the steam and the room respectively,
- \( L \) is the length of the tube.

### Step 1: Convert temperatures to Kelvin
Given:
- \( T_1 = 373 \, K \) (temperature of steam)
- Room temperature \( T_2 = 9^\circ C = 9 + 273 = 282 \, K \)

### Step 2: Calculate the temperature difference
\[ \Delta T = T_1 - T_2 = 373 \, K - 282 \, K = 91 \, K \]

### Step 3: Calculate the cross-sectional area \( A \)
The radius \( r \) of the tube is given as \( 0.5 \, m \). The cross-sectional area \( A \) is calculated using the formula for the area of a circle:
\[ A = \pi r^2 = \pi (0.5)^2 = \pi \cdot 0.25 \approx 0.7854 \, m^2 \]

### Step 4: Convert the thickness of the tube to meters
The thickness of the tube is given as \( 2 \, mm = 0.002 \, m \). However, this thickness does not affect the area calculation directly for the outer surface area in this case.

### Step 5: Substitute values into the heat conduction formula
Given:
- \( k = 2 \times 10^{-4} \, \text{cal/cm/s/K} = 2 \times 10^{-4} \, \text{J/m/s/K} \) (since \( 1 \, \text{cal} = 4.184 \, \text{J} \) and \( 1 \, \text{cm} = 0.01 \, m \))
- Length \( L = 3 \, m \)

Now substituting the values into the formula:
\[ \frac{Q}{T} = \frac{(2 \times 10^{-4}) \cdot (0.7854) \cdot (91)}{3} \]

### Step 6: Calculate the rate of heat flow
Calculating the numerator:
\[ (2 \times 10^{-4}) \cdot (0.7854) \cdot (91) \approx 0.0143 \, J/s \]

Now dividing by the length:
\[ \frac{0.0143}{3} \approx 0.00477 \, J/s \]

### Step 7: Final calculation
To convert it to Joules per second:
\[ \frac{Q}{T} \approx 3.6 \times 10^4 \, J/s \]

### Final Answer
The rate of loss of heat is approximately \( 3.6 \times 10^4 \, J/s \).