Test Index

Thermodynamics Part 1

Section: Physics

© examsnet.com

Question : 79 of 100

Marks: +1, -0

The figure shows a system of two concentric spheres of radii

r_1

r_2

are kept at temperatures

T_1

T_2

, respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to

$\ln \left( \frac{r_2}{r_1} \right)$
$\frac{r_2 - r_1}{r_1 r_2}$
$(r_2 - r_1)$
$\frac{r_1 r_2}{r_2 - r_1}$

Solution:

Consider a shell of thickness

(d r)

and of radius

(r)

and the temperature of inner and outer surfaces of this shell be

T,(T-d T)

\;H=\;\frac{K A[(T-d T)-T]}{d r}=\;\frac{-K A d T}{d r}

\;H=-4 \pi K r^2 \;\frac{d T}{d r}

as

A=4 \pi r^2)

Then,

(H) \int\limits_{r_1}^{r^2} \frac{1}{r^2} dr = -4 \pi K \int\limits_{T_1}^{T_2} dT

H \left[ \frac{1}{r_1} - \frac{1}{r_2} \right] = -4 \pi K [T_2 - T_1]

or

H = \frac{\overset{\rightarrow r_1}{-4 \pi K r_2 (T_2-T_1)}}{r_2 - r_1}

© examsnet.com

Go to Question:

Prev Question

Next Question