The thermal conductivity of a material in CGS system is 0.4. In steady state, the rate of flow of heat 10 cal/sec-cm, then the thermal gradient will be

Answer

To find the thermal gradient (dT/dx) given the thermal conductivity and the rate of heat flow, we can use Fourier's law of heat conduction, which states:

\[ Q = k \cdot A \cdot \frac{dT}{dx} \]

Where:
- \( Q \) is the rate of heat flow (in cal/sec),
- \( k \) is the thermal conductivity (in cal/sec-cm-°C),
- \( A \) is the area through which heat is flowing (in cm²),
- \( \frac{dT}{dx} \) is the thermal gradient (in °C/cm).

In this case, we are given:
- \( k = 0.4 \, \text{cal/sec-cm-°C} \)
- \( Q = 10 \, \text{cal/sec-cm} \)

Since the area \( A \) is not specified, we can assume it to be 1 cm² for simplicity. Thus, the equation simplifies to:

\[ Q = k \cdot \frac{dT}{dx} \]

Now, we can rearrange this equation to solve for the thermal gradient \( \frac{dT}{dx} \):

\[ \frac{dT}{dx} = \frac{Q}{k} \]

Substituting the known values into the equation:

\[ \frac{dT}{dx} = \frac{10 \, \text{cal/sec-cm}}{0.4 \, \text{cal/sec-cm-°C}} \]

Calculating this gives:

\[ \frac{dT}{dx} = \frac{10}{0.4} = 25 \, \text{°C/cm} \]

Thus, the thermal gradient \( \frac{dT}{dx} \) is:

**Final Answer: 25 °C/cm**

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