The figure shows a system of two concentric spheres of radii r1 and r2 and kept at temperatures T1 a

Q: Why can't we just use the average area of the two spheres?

Thermal resistance depends inversely on the area ($R \propto 1/A$). The average of the inverse is not equal to the inverse of the average. Because the area changes quadratically with radius, calculus is strictly required to sum the resistances correctly.

Q: Why are the concentric shells considered to be in series?

Heat flows radially from the inner sphere to the outer sphere. It must pass through every single concentric layer one after another. When heat (or current) passes through elements sequentially without branching, they are in series.

Q: What if the shape was a cylinder instead of a sphere?

For a cylinder, the cross-sectional area is $2\pi x L$, which depends linearly on $x$. Integrating $1/x$ gives a natural logarithm ($\ln$), which is why cylindrical heat flow involves logarithmic terms, unlike the $1/x^2$ dependence in spheres.

Q: Does it matter if $T_1 > T_2$ or $T_2 > T_1$?

Not for the proportionality! The direction of heat flow will reverse, changing the sign of the heat current $H$, but the geometric factor $\frac{r_1 r_2}{r_2 - r_1}$ remains exactly the same.

Q: Why do we use $dx$ as the length in the thermal resistance formula?

In the formula $R = \frac{l}{KA}$, '$l$' is the distance the heat travels through the material. For our infinitesimally thin shell, the heat travels radially across its thickness, which is exactly $dx$.

Question

The figure shows a system of two concentric spheres of radii $r_1$ and $r_2$ and kept at temperatures $T_1$ and $T_2$, respectively. The radial rate of flow of heat in a substance between the two concentric spheres, is proportional to

Options:

(A) $\frac{(r_2 - r_1)}{(r_1 r_2)}$

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

(C) $\frac{r_1 r_2}{(r_2 - r_1)}$

(D) $(r_2 - r_1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine two hollow spheres, one placed perfectly inside the other. The inner sphere has a radius $r_1$ and the outer has a radius $r_2$. They are maintained at different temperatures, $T_1$ and $T_2$. Because of this temperature difference, heat will naturally flow radially between them.

But there is a catch here. If you think about heat flowing through a standard uniform rod, the cross-sectional area remains constant. However, in our spherical geometry, as the heat travels outward from the center, the surface area it crosses is continuously expanding!

### The Master Equation and Calculus

Because the cross-sectional area is not constant, we cannot simply use our standard thermal resistance formula $R = \frac{l}{KA}$ directly for the entire gap. We need to bring in the power of calculus.

Let's consider an infinitely thin spherical shell at a distance $x$ from the center, with a tiny thickness $dx$. For this extremely thin shell, the surface area is practically constant, equal to the surface area of a sphere: $A = 4\pi x^2$.

So, its tiny thermal resistance, $dR$, is simply its thickness $dx$ divided by the thermal conductivity $K$ and its area:

$$dR = \frac{dx}{K \cdot 4\pi x^2}$$

### Integrating for Total Resistance

To find the total thermal resistance offered by the entire material between the two spheres, we must add up the resistances of all such thin concentric shells. This means we need to integrate $dR$ from the inner radius $r_1$ to the outer radius $r_2$.

$$R = \int_{r_1}^{r_2} \frac{dx}{4\pi K x^2}$$

Let's evaluate the integral. The integral of $\frac{1}{x^2}$ is $-\frac{1}{x}$. Applying the upper limit $r_2$ and lower limit $r_1$, we obtain the total thermal resistance $R$:

$$R = \frac{1}{4\pi K} \left[ -\frac{1}{x} \right]_{r_1}^{r_2} = \frac{1}{4\pi K} \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$$

Simplifying the fractions inside the bracket gives us:

$$R = \frac{r_2 - r_1}{4\pi K r_1 r_2}$$

### Final Calculation for Heat Current

Now, we need to find the radial rate of heat flow, or heat current $H$. Just like Ohm's law in electricity ($I = \frac{\Delta V}{R}$), heat current is the temperature difference divided by the total thermal resistance.

$$H = \frac{T_1 - T_2}{R}$$

Let's substitute the expression for $R$ that we just derived:

$$H = \frac{T_1 - T_2}{\frac{r_2 - r_1}{4\pi K r_1 r_2}}$$

Rearranging the terms, we get the final relation:

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

We can clearly see that the heat current is directly proportional to the product of the radii, and inversely proportional to their difference.

**$$H \propto \frac{r_1 r_2}{r_2 - r_1}$$**

This perfectly matches option (c). A beautiful application of calculus in physics!

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Visualized Solution

$P r o b l e m S e t u p$

$F o u r i e r^{'} s L a w o f H e a t C o n d u c t i o n$

$E l e m e n t a l T h er ma l R es i s t an ce$

$T o t a l T h er ma l R es i s t an ce$

$E v a l u a t in g t h e I n t e g r a l$

$R a d ia l R a t e o f H e a t F l o w$

$P r o p or t i o na l i t y$

The Sigma Insight: Heat Transfer

Analyzing the Setup

The Master Equation and Calculus

Integrating for Total Resistance

Final Calculation for Heat Current

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Problem Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fourier′s Law of Heat Conduction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Elemental Thermal Resistance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Thermal Resistance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating the Integral

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Radial Rate of Heat Flow

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Proportionality