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Question: Heat flows radially outward through a spherical shell of outside radius \({R}_{2}\) and inner radius...

Heat flows radially outward through a spherical shell of outside radius R2{R}_{2} and inner radius R1{R}_{1}, The temperature of inner radius is Θ1{ \Theta }_{ 1 } and that outer is Θ2{ \Theta }_{ 2 }. At what radial distance from center of shell the temperature is just half way between Θ1{ \Theta }_{ 1 } and Θ2{ \Theta }_{ 2}
A.R1+R22B.4R1+3R22C.8R2R14R1+3R2D.2R1R2R1+R2A. \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}}

Explanation

Solution

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 Θ1{ \Theta }_{ 1 } and Θ2{ \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=kAΔΘΔRH=\dfrac { kA\Delta \Theta }{ \Delta R }

Complete answer:
Given: Outer radius =R2{R}_{2}
Inner radius= R1{R}_{1}
Temperature of inner radius= Θ1{\Theta}_{1}
Temperature of the outer radius= Θ2{\Theta}_{2}
The rate of flow of heat through a conductor is given by,
H=kAΔΘΔRH=\dfrac { kA\Delta \Theta }{ \Delta R } …(1)
Where, k is the thermal conductivity
A is the cross-sectional area
ΔΘ\Delta \Theta is the change in the temperature
ΔR\Delta R is the change in the radius of the shell
Using the given data equation. (1) can be written as,
H=KA(Θ1−Θ2)R2−R1H=\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=kA((Θ1+Θ22)−Θ2)R2−xH= \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,
KA(Θ1−Θ2)R2−R1=kA((Θ1+Θ22)−Θ2)R2−x\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,
(Θ1−Θ2)R2−R1=(Θ1+Θ22)−Θ2R2−x\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 }
⇒(Θ1−Θ2)R2−R1=(Θ1+Θ2−2Θ22)R2−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 }
⇒(Θ1−Θ2)R2−R1=(Θ1−Θ22)R2−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 }
⇒(Θ1−Θ2)R2−R1=Θ1−Θ22(R−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.
1R2−R1=12(R−x)\dfrac { 1 }{ { R }_{ 2 }-{ R }_{ 1 } } =\dfrac { 1 }{ 2\left( R-x \right) }
⇒2(R−x)=R2−R1\Rightarrow 2\left( R-x \right) ={ R }_{ 2 }-{ R }_{ 1 }
⇒R−x=R2−R12\Rightarrow R-x= \dfrac {{ R }_{ 2 }-{ R }_{ 1 }}{2}
⇒x=R−R2−R12\Rightarrow x= R- \dfrac {{ R }_{ 2 }-{ R }_{ 1 }}{2}
∴x=R1+R22\therefore x= \dfrac {{R}_{1}+{R}_{2}}{2}
Thus, the temperature is just half way between Θ1{ \Theta }_{ 1 } and Θ2{ \Theta }_{ 2} at R1+R22\dfrac {{R}_{1}+{R}_{2}}{2}.

So, the correct answer is option A i.e. R1+R22\dfrac {{R}_{1}+{R}_{2}}{2}.

Note:
We know, the flow of current can be written as,
I=V2−V1RI= \dfrac {{V}_{2}-{V}_{1}}{R} …(1)
Where, V2{V}_{2} is the final potential difference
V1{V}_{1} is the initial potential difference
R is the resistance
The formula for resistance is given by,
R=ρlAR=\dfrac { \rho l }{ A } …(2)
Where, A is the cross-sectional area
l is the length
ρ\rho is the resistivity
Substituting equation. (2) in equation. (1) we get,
I=V2−V1ρlAI= \dfrac { { V }_{ 2 }-{ V }_{ 1 } }{ \dfrac { \rho l }{ A } }
⇒I=A(V2−V1)ρl\Rightarrow I= \dfrac {A\left({V}_{2}-{V}_{1}\right)}{\rho l}
This same current will flow through all the points across the shell.