Question 674: Thermodynamics & KTG

Question

The two ends of a rod of length ' $x$ ' and uniform cross-sectional area ' A ' are kept at temperatures ' $\mathrm{T}_1$ ' and ' $\mathrm{T}_2$ ' respectively ( $\mathrm{T}_1>\mathrm{T}_2$ ). If the rate of heat transfer is ' $\mathrm{Q} / \mathrm{t}$ ', through the rod in steady state, then the coefficient of thermal conductivity ' K ' is

Accepted Answer

We are asked about heat conduction through a rod.
The formula for the rate of heat conduction is FourierÃ¢ÂÂs law:
$ \frac{Q}{t} \;=\; \frac{K A (T_1 - T_2)}{x} $
Where
$K$ = thermal conductivity
$A$ = cross-sectional area
$x$ = length of the rod
$(T_1 - T_2)$ = temperature difference
$\frac{Q}{t}$ = rate of heat flow
Now we rearrange for $K$:
$ K = \frac{	frac{Q}{t} \cdot x}{A (T_1 - T_2)} $
$ K = \frac{x Q}{t A (T_1 - T_2)} $
Ã¢ÂÂ Correct option is:
Option B:
$ \frac{x Q}{t A \,(T_1 - T_2)} $

Answer

$\frac{A Q}{\operatorname{tx}\left(T_1-T_2\right)}$

Answer

$\frac{x Q}{t A\left(T_1-T_2\right)}$

Answer

$\frac{x A Q}{t\left(T_1-T_2\right)}$

Answer

$\frac{Q}{\operatorname{txA}\left(T_1-T_2\right)}$

Thermodynamics & KTG

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