Name: Cause of Convection
Uploaded: 2019-07-19
Duration: 2 min 7 s
Description: In this video, we analyse the process of Convection and understand how important it is for convectional rainfall, mantle convection currents and many other t...

Question

Define &lsquo;coefficient of thermal conductivity&rsquo; and state its SI unit and dimensions.

Accepted Answer

Let us consider a parallel-faced slab whose cross-sectional area is $A$ and length $l$ . Let its faces be maintained at steady temperatures $θ_{1}$ and $θ_{2}$ . Heat will flow through the slab from the face at higher temperature $θ_{1}$ to that at the lower temperature $θ_{2}$ .

Let us assume that no heat escapes out from the sides of the slab by convection or radiation. Thus, the lines of flow will be at all points normal to the faces of the slab. In these circumstances, under steady state, the quantity of heat $Q$ entering one face in a given time is same as the heat leaving the other face in the same time. Experiment shows that the heat $Q$ is -

$1 .$ Directly proportional to the cross-sectional area $A$ of each face, that is $Q \propto A$ .

$2 .$ Directly proportional to the temperature-difference $(θ_{1} - θ_{2})$ , that is $Q \propto (θ_{1} - θ_{2})$ .

$3 .$ Directly proportional to the time $t$ for which the heat flows, that is, $Q \propto t$ .

$4 .$ Inversely proportional to the length $l$ of the slab (distance between hot and cold faces), that is, $Q \propto \frac{1}{l}$ .

Combining the above factors, we get,

$Q \propto A \frac{(θ_{1} - θ_{2}) t}{l}$ or

$Q = K A \frac{(θ_{1} - θ_{2}) t}{l}$ , where $K$ is a constant depending on the material of the slab. It is called the “coefficient of thermal conductivity” of the material of the slab. The coefficient of thermal conductivity of a material is defined as the amount of heat that flows in unit time through unit area of the material perpendicular to the flow under unit temperature-gradient, when the steady state has been reached.

The SI unit of $K$ is ${Js}^{- 1} m^{- 1} ° C^{- 1} or {Js}^{- 1} m^{- 1} K^{- 1}$ .

The dimensional formula for $K$ is, $[{MLT}^{- 3} θ^{- 1}]$ .

G L Mittal and TARUN MITTAL Solutions for Chapter: Thermal Conduction, Exercise 1: QUESTIONS

G L Mittal Physics Solutions for Exercise - G L Mittal and TARUN MITTAL Solutions for Chapter: Thermal Conduction, Exercise 1: QUESTIONS

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