One end of a rod length 20cm is inserted in a furnace at 800K. The sides of the rod are covered with an insulating material and the other end emits radiation like a blackbody. The temperature of this end is 750K in the steady state. The temperature of the surrounding air is 300K. Assuming radiation to be the only important mode of energy transfer between the surrounding and the open end of the rod, find the thermal conductivity of the rod. Stefan constant `sigma=6.0xx10^(-1)Wm^(-2)K^(-4)` .

Explanation

To solve the problem, we need to find the thermal conductivity of the rod given the conditions of heat transfer through radiation. Here’s a step-by-step solution:

### Step 1: Understand the Heat Transfer Mechanism
The rod is in a furnace at 800 K, and one end of the rod is at 750 K, while the surrounding air is at 300 K. The only mode of heat transfer from the open end of the rod to the surroundings is through radiation.

### Step 2: Set Up the Heat Transfer Equations
In the steady state, the heat conducted through the rod equals the heat radiated from the open end.

1. **Heat Conducted (Q_conducted)**:
\[ Q_{\text{conducted}} = K \cdot A \cdot \frac{(T_f - T_1)}{L} \]
where:
- \( K \) = thermal conductivity of the rod
- \( A \) = cross-sectional area of the rod (we will see it cancels out)
- \( T_f \) = temperature of the furnace = 800 K
- \( T_1 \) = temperature of the open end = 750 K
- \( L \) = length of the rod = 20 cm = 0.2 m

2. **Heat Radiated (Q_radiated)**:
According to the Stefan-Boltzmann law:
\[ Q_{\text{radiated}} = \epsilon \cdot \sigma \cdot A \cdot (T_1^4 - T_s^4) \]
where:
- \( \epsilon \) = emissivity (for a black body, \( \epsilon = 1 \))
- \( \sigma \) = Stefan-Boltzmann constant = \( 6.0 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \)
- \( T_s \) = temperature of the surroundings = 300 K

### Step 3: Equate the Heat Conducted and Heat Radiated
Setting the two equations equal:
\[ K \cdot A \cdot \frac{(800 - 750)}{0.2} = \epsilon \cdot \sigma \cdot A \cdot (750^4 - 300^4) \]

### Step 4: Simplify the Equation
Since the area \( A \) appears on both sides, we can cancel it out:
\[ K \cdot \frac{(800 - 750)}{0.2} = \sigma \cdot (750^4 - 300^4) \]

### Step 5: Calculate the Values
1. Calculate the left side:
\[ K \cdot \frac{50}{0.2} = 250K \]

2. Calculate the right side:
- First, calculate \( 750^4 \) and \( 300^4 \):
\[ 750^4 = 316406250000 \quad \text{and} \quad 300^4 = 8100000000 \]
- Now calculate \( 750^4 - 300^4 \):
\[ 750^4 - 300^4 = 316406250000 - 8100000000 = 308306250000 \]
- Now substitute into the equation:
\[ K \cdot 250 = 6.0 \times 10^{-8} \cdot 308306250000 \]

### Step 6: Solve for K
\[ K \cdot 250 = 6.0 \times 10^{-8} \cdot 308306250000 \]
Calculating the right side:
\[ K \cdot 250 = 18498.375 \]
Now solve for \( K \):
\[ K = \frac{18498.375}{250} = 73.9935 \approx 74 \, \text{W/m°C} \]

### Final Answer
The thermal conductivity of the rod is approximately **74 W/m°C**.