Law Of Thermal Conductivity - Practice Questions & MCQ

Law of Thermal Conductivity - 

Consider a rod of length 'l' , area of cross-section 'A' whose faces are maintained at temperature and respectively. In steady state the amount of heat flowing from one face to the other face in time t is given by - 

$
\begin{aligned}
& Q=\frac{K A\left(\theta_1-\theta_2\right) t}{l} \\
& Q=\text { Amount of heat transfer } \\
& t=\text { Time of heat flow } \\
& K=\text { Thermal conductivity of the material }
\end{aligned}
$

So, from the above equation we can calculate the - Rate of flow of heat i.e. heat current which can be written as -

$
\frac{Q}{t}=H=\frac{K A\left(\theta_1-\theta_2\right)}{l}
$

In the differential form, this heat current can also be written as -

$
\frac{d Q}{d t}=-K A \frac{d \theta}{d x}
$

In case of non-steady state or variable cross-section, this is the more general equation can be used to solve problems.

Relation of thermal conductivity of some material -

$
\begin{aligned}
& K_{A g}>K_{C u}>K_{A l} \\
& K_{\text {Solid }}>K_{\text {Liquid }}>K_{\text {Gas }} \\
& K_{\text {Metals }}>K_{\text {Non-metals }}
\end{aligned}
$

Thermal resistance $\left(R_{t h}\right)$ : The thermal resistance of a body is defined as the measure of its opposition to the flow of heat through it. It is defined as the ratio of temperature difference to the heat current $\left(\frac{Q}{t}\right)$

$
R_{t h}=\frac{\theta_1-\theta_2}{H}=\frac{\theta_1-\theta_2}{K A\left(\theta_1-\theta_2\right) / l}=\frac{l}{K A}
$