If the coefficient of conductivity of aluminium is `0.5 cal cm^(-1) s^(-1).^(@)C^(-1)`, then the other to conductor `10 cal s^(-1) cm^(-2)` in the steady state, the temperature gradient in aluminium must be

Options

A. `5^(@)C//cm`
B. `10^(@)C//cm`
C. `20^(@)C//cm`
D. `10.5^(@)C//cm`

Explanation

To find the temperature gradient in aluminum, we can use the formula for heat conduction, which is given by Fourier's law of heat conduction:

\[ \frac{DQ}{DT} = K \cdot A \cdot \frac{dT}{dx} \]

Where:
- \(\frac{DQ}{DT}\) is the heat transfer rate (thermal current),
- \(K\) is the thermal conductivity,
- \(A\) is the cross-sectional area,
- \(\frac{dT}{dx}\) is the temperature gradient.

Given:
- The thermal conductivity of aluminum, \(K = 0.5 \, \text{cal} \, \text{cm}^{-1} \, \text{s}^{-1} \, \text{°C}^{-1}\)
- The heat transfer rate through the other conductor, \(\frac{DQ}{DT} = 10 \, \text{cal} \, \text{s}^{-1} \, \text{cm}^{-2}\)

Since we are in a steady state, the heat transfer rate through aluminum must equal the heat transfer rate through the other conductor. Therefore, we can set:

\[ \frac{DQ}{DT} = K \cdot \frac{dT}{dx} \]

Substituting the known values into the equation, we have:

\[ 10 = 0.5 \cdot \frac{dT}{dx} \]

Now, we need to solve for \(\frac{dT}{dx}\):

1. Rearranging the equation gives:

\[ \frac{dT}{dx} = \frac{10}{0.5} \]

2. Performing the division:

\[ \frac{dT}{dx} = 20 \, \text{°C/cm} \]

Thus, the temperature gradient in aluminum must be \(20 \, \text{°C/cm}\).