Question

The thickness of a metallic plate is 0.4cm. The temperature between its two surfaces is \[{20^0}C\] . The quantity of heat flowing per second is 50 calories from \[5c{m^2}\]area. Find the coefficient of thermal conductivity (in CGS system).
A. 0.2
B. 0.3
C. 0.4
D. 0.5

Answer 1

Hint:Before we proceed with the problem, it is important to understand the coefficient of thermal conductivity. It is defined as the capability of an object or a material to transfer heat from the hot side to the cold side. The coefficient of thermal conductivity is denoted by K and its unit is watts per meter per kelvin.

Formula Used:
To find the rate of flow of heat the formula is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Where, A is cross-sectional area, \[\Delta T\] is the temperature difference between two ends, L is the length of the cylinder and K is the thermal conductivity.

Complete step by step solution:
Consider a metallic plate of thickness \[L = 0.4\,cm\] and the temperature between its two surfaces is \[\Delta T = {20^0}C\] . The quantity of heat flowing per second is 50 calories that is \[\dfrac{Q}{t} = 50\] from the area \[A = 5\,c{m^2}\]. We need to find the coefficient of thermal conductivity.

We know that to find the rate of flow of heat the formula is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Substitute all the given values then we obtain,
\[50 = K \times \dfrac{{5 \times 20}}{{0.4}}\]
\[\Rightarrow K = \dfrac{{50 \times 0.4}}{{5 \times 20}}\]
\[\therefore K = 0.2\]
Therefore, the coefficient of thermal conductivity is 0.2.

Hence, option A is the correct answer.

Note:The thermal conductivity is the properties of a solid. When we talk about fluids, the convective heat transfer coefficient (CHTC) defines the heat transfer. It is also dependent on the velocity of fluid flow.