Heat And Thermodynamics Ques 10
10 A heat source at $T=10^3 \mathrm{~K}$ is connected to another heat reservoir at $T=10^2 \mathrm{~K}$ by a copper slab which is $1 \mathrm{~m}$ thick. Given that the thermal conductivity of copper is $0.1 \mathrm{WK}^{-1} \mathrm{~m}^{-1}$, the energy flux through it in the steady state is
(2019 Main, 10 Jan I)
(a) $90 \mathrm{Wm}^{-2}$
(b) $65 \mathrm{Wm}^{-2}$
(c) $120 \mathrm{Wm}^{-2}$
(d) $200 \mathrm{Wm}^{-2}$
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Answer:
Correct Answer: 10.( a )
Solution:
Formula:
- Energy flux is the rate of heat flow per unit area through the rod.
Also, rate of flow of heat per unit time through a material of area of cross-section $A$ and thermal conductivity $k$ between the temperatures $T_1$ and $T_2\left(T_1>T_2\right)$ is given as,
$ \frac{\Delta Q}{\Delta t}=\frac{k A\left(T_1-T_2\right)}{l} $
Energy flux using Eq. (i), we get
$ =\frac{1}{A} \cdot \frac{\Delta Q}{\Delta t}=\frac{k\left(T_1-T_2\right)}{l} $
Here, $k=0.1 \mathrm{~W} \mathrm{~K}^{-1} \mathrm{~m}^{-1}, l=1 \mathrm{~m}$
$ T_1=1000 \mathrm{~K} \text { and } T_2=100 \mathrm{~K} $
$ \begin{aligned} \therefore \quad \text { Energy flux } & =\frac{01(1000-100)}{1} \\ & =90 \mathrm{Wm}^{-2} . \end{aligned} $
BETA
