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

Options

A. `10^(@)C//cm`
B. `12^(@)C//cm`
C. `25^(@)C//cm`
D. `20^(@)C//cm`

Answer

To solve the problem, we will use the formula for the rate of heat flow through a material, which is given by Fourier's law of heat conduction:

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

where:
- \(\frac{dQ}{dt}\) = rate of heat flow (in cal/sec)
- \(k\) = thermal conductivity (in cal/sec-cm-°C)
- \(A\) = cross-sectional area (in cm²)
- \(\frac{dT}{dx}\) = thermal gradient (in °C/cm)

### Step 1: Identify the given values
From the problem, we have:
- Thermal conductivity, \(k = 0.4 \, \text{cal/sec-cm-°C}\)
- Rate of heat flow, \(\frac{dQ}{dt} = 10 \, \text{cal/sec}\)
- We will assume the area \(A = 1 \, \text{cm}^2\) since it is not specified.

### Step 2: Rearrange the formula to find the thermal gradient
We need to find \(\frac{dT}{dx}\), so we rearrange the formula:

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

### Step 3: Substitute the known values into the equation
Substituting the known values into the rearranged formula:

\[ \frac{dT}{dx} = \frac{10 \, \text{cal/sec}}{0.4 \, \text{cal/sec-cm-°C} \cdot 1 \, \text{cm}^2} \]

### Step 4: Calculate the thermal gradient
Now we perform the calculation:

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

### Final Result
The thermal gradient \(\frac{dT}{dx}\) is \(25 \, \text{°C/cm}\).

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