The heat is flowing through a rod of length 50 cm and area of cross-section `5cm^(2)` . Its ends are respectively at `25^(@)C` and `125^(@)C` . The coefficient of thermal conductivity of the material of the rod is `0.092 kcal // m × s ×.^(@) C` . The temperature gradient in the rod is

Options

A. `2^(@)C//cm`
B. `2^(@)C//m`
C. `20^(@)C//m`
D. `20^(@)C//cm`

Answer

To find the temperature gradient in the rod, we can follow these steps:

### Step 1: Identify the given values
- Length of the rod (L) = 50 cm = 0.5 m (since we will convert to meters for standard units)
- Temperature at one end (T1) = 25°C
- Temperature at the other end (T2) = 125°C
- Change in temperature (ΔT) = T2 - T1 = 125°C - 25°C = 100°C
- Length of the rod in meters (Δx) = 0.5 m

### Step 2: Calculate the temperature gradient
The temperature gradient (∇T) is defined as the change in temperature per unit length. It can be calculated using the formula:

\[ \text{Temperature Gradient} (\nabla T) = \frac{\Delta T}{\Delta x} \]

Substituting the values we have:

\[ \nabla T = \frac{100°C}{0.5 m} = 200°C/m \]

### Step 3: Convert the temperature gradient to cm
Since the original length was given in centimeters, we can convert the temperature gradient to °C/cm:

\[ \nabla T = 200°C/m = 20°C/10cm = 20°C/cm \]

### Final Answer
So, the temperature gradient in the rod is:

\[ \nabla T = 200°C/m \text{ or } 20°C/cm \]

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