A steel frame `(K=45Wm^(-1)C^(-1))` of total length `60cm` and cross sectional area `0.2cm^(2)` , Forms three side of a square. The free ends are maintained at `20^(@)C` and `40^(@)C` . Find the rate of heat flow through a cross section of the frame.

Explanation

To find the rate of heat flow through the cross-section of the steel frame, we can use the formula for heat conduction, which is given by Fourier's law:

\[ Q = \frac{K \cdot A \cdot \Delta T}{L} \]

Where:
- \( Q \) = rate of heat flow (in watts)
- \( K \) = thermal conductivity of the material (in W/m°C)
- \( A \) = cross-sectional area (in m²)
- \( \Delta T \) = temperature difference (in °C)
- \( L \) = length of the material (in meters)

### Step 1: Convert the given values to appropriate units
1. **Thermal conductivity \( K \)**: Given as \( 45 \, \text{W/m°C} \) (no conversion needed).
2. **Length \( L \)**: Given as \( 60 \, \text{cm} \). Convert to meters:
\[ L = 60 \, \text{cm} = 0.60 \, \text{m} \]
3. **Cross-sectional area \( A \)**: Given as \( 0.2 \, \text{cm}^2 \). Convert to square meters:
\[ A = 0.2 \, \text{cm}^2 = 0.2 \times 10^{-4} \, \text{m}^2 = 2 \times 10^{-5} \, \text{m}^2 \]
4. **Temperature difference \( \Delta T \)**: The temperatures are \( 40°C \) and \( 20°C \):
\[ \Delta T = 40°C - 20°C = 20°C \]

### Step 2: Substitute the values into the formula
Now we can substitute the values into the formula:
\[ Q = \frac{K \cdot A \cdot \Delta T}{L} \]
\[ Q = \frac{45 \, \text{W/m°C} \cdot 2 \times 10^{-5} \, \text{m}^2 \cdot 20 \, \text{°C}}{0.60 \, \text{m}} \]

### Step 3: Calculate the rate of heat flow
Calculating the numerator:
\[ Q = \frac{45 \cdot 2 \times 10^{-5} \cdot 20}{0.60} \]
\[ Q = \frac{1800 \times 10^{-5}}{0.60} \]
\[ Q = \frac{0.018}{0.60} = 0.03 \, \text{W} \]

### Final Answer
The rate of heat flow through the cross-section of the frame is:
\[ Q = 0.03 \, \text{W} \]