Question

Heat flows radially outward through a spherical shell of outside radius ${R}_{2}$ and inner radius ${R}_{1}$, The temperature of inner radius is ${ \Theta }_{ 1 }$ and that outer is ${ \Theta }_{ 2 }$. At what radial distance from center of shell the temperature is just half way between ${ \Theta }_{ 1 }$ and ${ \Theta }_{ 2}$
$A. \dfrac {{R}_{1}+{R}_{2}}{2}
B. \dfrac {4{R}_{1}+3{R}_{2}}{2}
C. \dfrac {8{R}_{2}{R}_{1}}{4{R}_{1}+3{R}_{2}}
D. \dfrac {2{R}_{1}{R}_{2}}{{R}_{1}+{R}_{2}}$

Answer 1

Hint: To solve this problem, use the formula for rate of flow of heat through a conductor. Substitute the values and obtain an equation. Then, assume at distance x the temperature is just half way between ${ \Theta }_{ 1 }$ and ${ \Theta }_{ 2}$. Obtain an expression for the rate of flow of heat at this distance x. Now, compare both the equations for rate of heat flow. Evaluate the obtained expression and solve it to find the value of x.

Formula used:
$H=\dfrac { kA\Delta \Theta }{ \Delta R }$

Complete answer:
Given: Outer radius =${R}_{2}$
            Inner radius= ${R}_{1}$
            Temperature of inner radius= ${\Theta}_{1}$
            Temperature of the outer radius= ${\Theta}_{2}$
The rate of flow of heat through a conductor is given by,
$H=\dfrac { kA\Delta \Theta }{ \Delta R }$ …(1)
Where, k is the thermal conductivity
            A is the cross-sectional area
            $\Delta \Theta$ is the change in the temperature
            $\Delta R$ is the change in the radius of the shell
Using the given data equation. (1) can be written as,
$H=\dfrac { KA\left( { \Theta }_{ 1 }-{ \Theta }_{ 2 } \right) }{ { R }_{ 2 }-{ R }_{ 1 } }$ …(2)
Now, let x be the distance from the center where we have to find the temperature.
So, the rate of heat flow at that point will be given by,
$H= \dfrac { kA\left( \left( \dfrac { { \Theta }_{ 1 }+{ \Theta }_{ 2 } }{ 2 } \right) -{ \Theta }_{ 2 } \right) }{ { R }_{ 2 }-x }$ …(3)
Comparing equation. (2) and (3) we get,
$\dfrac { KA\left( { \Theta }_{ 1 }-{ \Theta }_{ 2 } \right) }{ { R }_{ 2 }-{ R }_{ 1 } }=\dfrac { kA\left( \left( \dfrac { { \Theta }_{ 1 }+{ \Theta }_{ 2 } }{ 2 } \right) -{ \Theta }_{ 2 } \right) }{ { R }_{ 2 }-x }$
Cancelling the common term on both the side we get,
$\dfrac { \left( { \Theta }_{ 1 }-{ \Theta }_{ 2 } \right) }{ { R }_{ 2 }-{ R }_{ 1 } }=\dfrac { \left( \dfrac { { \Theta }_{ 1 }+{ \Theta }_{ 2 } }{ 2 } \right) -{ \Theta }_{ 2 } }{ { R }_{ 2 }-x }$
$\Rightarrow \dfrac { \left( { \Theta }_{ 1 }-{ \Theta }_{ 2 } \right) }{ { R }_{ 2 }-{ R }_{ 1 } }=\dfrac { \left( \dfrac { { \Theta }_{ 1 }+{ \Theta }_{ 2 }-2{ \Theta }_{ 2 } }{ 2 } \right) }{ { R }_{ 2 }-x }$
$\Rightarrow \dfrac { \left( { \Theta }_{ 1 }-{ \Theta }_{ 2 } \right) }{ { R }_{ 2 }-{ R }_{ 1 } }=\dfrac { \left( \dfrac { { \Theta }_{ 1 }-{ \Theta }_{ 2 } }{ 2 } \right) }{ { R }_{ 2 }-x }$
$\Rightarrow \dfrac { \left( { \Theta }_{ 1 }-{ \Theta }_{ 2 } \right) }{ { R }_{ 2 }-{ R }_{ 1 }}= \dfrac { { \Theta }_{ 1 }-{ \Theta }_{ 2 } }{ 2\left( R-x \right) }$
Numerator on both sides are same. So, we can cancel them.
$ \dfrac { 1 }{ { R }_{ 2 }-{ R }_{ 1 } } =\dfrac { 1 }{ 2\left( R-x \right) } $
$\Rightarrow 2\left( R-x \right) ={ R }_{ 2 }-{ R }_{ 1 }$
$\Rightarrow R-x= \dfrac {{ R }_{ 2 }-{ R }_{ 1 }}{2}$
$\Rightarrow x= R- \dfrac {{ R }_{ 2 }-{ R }_{ 1 }}{2}$
$\therefore x= \dfrac {{R}_{1}+{R}_{2}}{2}$
Thus, the temperature is just half way between ${ \Theta }_{ 1 }$ and ${ \Theta }_{ 2}$ at $\dfrac {{R}_{1}+{R}_{2}}{2}$.

So, the correct answer is option A i.e. $\dfrac {{R}_{1}+{R}_{2}}{2}$.

Note:
We know, the flow of current can be written as,
$I= \dfrac {{V}_{2}-{V}_{1}}{R}$ …(1)
Where, ${V}_{2}$ is the final potential difference
             ${V}_{1}$ is the initial potential difference
             R is the resistance
The formula for resistance is given by,
$R=\dfrac { \rho l }{ A }$ …(2)
Where, A is the cross-sectional area
             l is the length
             $\rho$ is the resistivity
Substituting equation. (2) in equation. (1) we get,
$I= \dfrac { { V }_{ 2 }-{ V }_{ 1 } }{ \dfrac { \rho l }{ A } }$
$\Rightarrow I= \dfrac {A\left({V}_{2}-{V}_{1}\right)}{\rho l}$
This same current will flow through all the points across the shell.