The ends of a copper rod of length 1m and area of cross-section `1cm^2` are maintained at `0^@C` and `100^@C`. At the centre of the rod there is a source of heat of power 25 W. Calculate the temperature gradient in the two halves of the rod in steady state. Thermal conductivity of copper is `400 Wm^-1 K^-1`.

Explanation

To solve the problem of finding the temperature gradient in the two halves of a copper rod with a heat source in the center, we will follow these steps:

### Step 1: Understand the setup
We have a copper rod of length \( L = 1 \, \text{m} \) and cross-sectional area \( A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2 \). The ends of the rod are maintained at temperatures \( T_1 = 0^\circ C \) and \( T_2 = 100^\circ C \). There is a heat source in the center of the rod that provides power \( P = 25 \, \text{W} \).

### Step 2: Calculate the total temperature difference
The total temperature difference across the rod is given by:
\[ \Delta T = T_2 - T_1 = 100^\circ C - 0^\circ C = 100 \, \text{K} \]

### Step 3: Calculate the heat transfer in steady state
In steady state, the heat transfer through the rod can be described by Fourier's law of heat conduction:
\[ P = k \cdot A \cdot \frac{\Delta T}{L} \]
where \( k \) is the thermal conductivity of copper, \( k = 400 \, \text{Wm}^{-1} \text{K}^{-1} \).

### Step 4: Calculate the temperature gradient
We can rearrange the formula to find the temperature gradient \( \frac{\Delta T}{L} \):
\[ \frac{\Delta T}{L} = \frac{P}{k \cdot A} \]
Substituting the known values:
\[ \frac{\Delta T}{L} = \frac{25 \, \text{W}}{400 \, \text{Wm}^{-1} \text{K}^{-1} \cdot 1 \times 10^{-4} \, \text{m}^2} \]

Calculating this gives:
\[ \frac{\Delta T}{L} = \frac{25}{400 \times 1 \times 10^{-4}} = \frac{25}{0.04} = 625 \, \text{K/m} \]

### Step 5: Determine the temperature gradient in each half
Since the heat source is at the center, the rod is divided into two halves. The temperature gradient in each half can be considered uniform. Therefore, the temperature gradient in each half of the rod is:
\[ \text{Temperature Gradient} = \frac{625 \, \text{K/m}}{2} = 312.5 \, \text{K/m} \]

### Final Answer
The temperature gradient in the two halves of the rod in steady state is \( 312.5 \, \text{K/m} \).

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