The thickness of a metallic plate is 0.4 cm. The temperature between its two surfaces is `20^(@)C`. The quantity of heat flowing per second is 50 calories from `5cm^(2)` area. In CGS system, the coefficient of thermal conductivity will be

Options

A. 0.2
B. 0.3
C. 0.4
D. 0.5

Explanation

To find the coefficient of thermal conductivity (k) of the metallic plate, we can use the formula for heat transfer through a material:

\[ Q = \frac{k \cdot A \cdot \Delta T}{d} \]

Where:
- \( Q \) = heat transfer per second (in calories)
- \( k \) = coefficient of thermal conductivity (in cal/cm·s·°C)
- \( A \) = area of the surface (in cm²)
- \( \Delta T \) = temperature difference (in °C)
- \( d \) = thickness of the material (in cm)

Given:
- \( Q = 50 \) calories
- \( A = 5 \) cm²
- \( \Delta T = 20 \) °C (the temperature difference is assumed to be 20 °C as the problem states the temperature between the two surfaces is 20 °C)
- \( d = 0.4 \) cm

Now, we can rearrange the formula to solve for \( k \):

\[ k = \frac{Q \cdot d}{A \cdot \Delta T} \]

Substituting the known values into the equation:

\[ k = \frac{50 \cdot 0.4}{5 \cdot 20} \]

Calculating the numerator:

\[ 50 \cdot 0.4 = 20 \]

Calculating the denominator:

\[ 5 \cdot 20 = 100 \]

Now substituting these values back into the equation for \( k \):

\[ k = \frac{20}{100} = 0.2 \]

Thus, the coefficient of thermal conductivity \( k \) is:

\[ \boxed{0.2} \text{ cal/cm·s·°C} \]