The left end of a copper rod (length=`20cm`, Area of cross section=`0.2cm^(2)`) is maintained at `20^(@)C` and the right end is maintained at `80^(@)C` . Neglecting any loss of heat through radiation, find (a) the temperature at a point 11cm from the left end and (b) the heat current through the rod. thermal conductivity of copper `=385Wm^(-1)C^(-1)` .

Explanation

To solve the problem, we will break it down into two parts: (a) finding the temperature at a point 11 cm from the left end of the rod, and (b) calculating the heat current through the rod.

### Given Data:
- Length of the rod, \( L = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Area of cross-section, \( A = 0.2 \, \text{cm}^2 = 0.2 \times 10^{-4} \, \text{m}^2 = 2 \times 10^{-5} \, \text{m}^2 \)
- Temperature at the left end, \( T_L = 20^\circ C \)
- Temperature at the right end, \( T_R = 80^\circ C \)
- Thermal conductivity of copper, \( k = 385 \, \text{W/m} \cdot \text{C} \)

### Part (a): Finding the Temperature at a Point 11 cm from the Left End

1. **Identify the position of the point**:
- The point is located at \( x = 11 \, \text{cm} = 0.11 \, \text{m} \) from the left end.

2. **Use the formula for temperature distribution**:
The temperature \( T(x) \) at a distance \( x \) from the left end can be found using the linear temperature gradient:
\[ T(x) = T_L + \left( \frac{T_R - T_L}{L} \right) x \]
Substituting the known values:
\[ T(0.11) = 20 + \left( \frac{80 - 20}{0.2} \right) \cdot 0.11 \]
\[ T(0.11) = 20 + \left( \frac{60}{0.2} \right) \cdot 0.11 \]
\[ T(0.11) = 20 + 300 \cdot 0.11 \]
\[ T(0.11) = 20 + 33 = 53^\circ C \]

**Temperature at a point 11 cm from the left end is \( 53^\circ C \).**

### Part (b): Calculating the Heat Current through the Rod

1. **Use the formula for heat current**:
The heat current \( I \) can be calculated using the formula:
\[ I = \frac{k \cdot A \cdot (T_R - T_L)}{L} \]
Substituting the known values:
\[ I = \frac{385 \cdot (2 \times 10^{-5}) \cdot (80 - 20)}{0.2} \]
\[ I = \frac{385 \cdot (2 \times 10^{-5}) \cdot 60}{0.2} \]
\[ I = \frac{385 \cdot 1.2 \times 10^{-3}}{0.2} \]
\[ I = \frac{0.462}{0.2} = 2.31 \, \text{W} \]

**The heat current through the rod is \( 2.31 \, \text{W} \).**

### Summary of Answers:
- (a) Temperature at a point 11 cm from the left end: \( 53^\circ C \)
- (b) Heat current through the rod: \( 2.31 \, \text{W} \)