Two thin metallic spherical shells of radii r1 and r2 (r1

Q: Why can't we just use the formula $R = \frac{L}{KA}$ directly?

The formula $R = \frac{L}{KA}$ assumes that the cross-sectional area $A$ is constant along the length $L$. In a spherical geometry, the area $A = 4\pi r^2$ changes continuously with the radius $r$. Therefore, we must use integration to sum up the resistances of infinitesimally thin shells.

Q: Why is the temperature difference written as $\theta_2 - \theta_1$ and not the other way around?

Heat always flows from a region of higher temperature to a region of lower temperature. The problem states that $\theta_2 > \theta_1$, so heat flows inwards. To keep the rate of heat flow positive, we take the magnitude of the difference, which is $\theta_2 - \theta_1$.

Q: What exactly is the area $A$ in the elemental resistance formula?

The area $A$ is the surface area of the imaginary spherical shell at radius $r$, which is $4\pi r^2$. This is the area perpendicular to the direction of the radial heat flow.

Q: How do we integrate $\frac{1}{r^2}$?

You can rewrite $\frac{1}{r^2}$ as $r^{-2}$. Using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1}$, the integral of $r^{-2}$ becomes $\frac{r^{-1}}{-1}$, which simplifies to $-\frac{1}{r}$.

Q: Does the direction of heat flow (inward vs outward) affect the total thermal resistance?

No, thermal resistance is an intrinsic property of the system's geometry and the material's thermal conductivity. It remains exactly the same whether heat flows from the inside out or the outside in.

Question

Two thin metallic spherical shells of radii $r_1$ and $r_2$ ($r_1 < r_2$) are placed with their centres coinciding. A material of thermal conductivity $K$ is filled in the space between the shells. The inner shell is maintained at temperature $	heta_1$ and the outer shell at temperature $	heta_2$ ($	heta_1 < 	heta_2$). The rate at which heat flows radially through the material is

Options:

(A) $\frac{4\pi K r_1 r_2 (	heta_2 - 	heta_1)}{r_2 - r_1}$

(B) $\frac{\pi r_1 r_2 (	heta_2 - 	heta_1)}{r_2 - r_1}$

(C) $\frac{K(	heta_2 - 	heta_1)}{r_2 - r_1}$

(D) $\frac{K(	heta_2 - 	heta_1)(r_2 - r_1)}{4\pi r_1 r_2}$

Sigma Percentile · Accepted Answer

### The Geometry of the Problem

Imagine you are standing inside a hollow metallic sphere, which is itself enclosed by a larger metallic sphere. The space between these two spheres is completely filled with a material that conducts heat. The inner sphere is kept at a cooler temperature $	heta_1$, while the outer sphere is blazing hot at a temperature $	heta_2$. Because nature abhors a temperature gradient, heat will relentlessly flow from the hotter outer shell to the cooler inner shell. Our mission is to find the exact rate at which this heat flows radially inwards.

### The Concept of Thermal Resistance

If this were a simple uniform rod, we could just use the standard formula for thermal resistance, $R = \frac{L}{KA}$. But here, we have a catch. As heat travels radially, the cross-sectional area it passes through isn't constant. The surface area of a sphere is $4\pi r^2$, which means the area shrinks as the heat moves inward from $r_2$ to $r_1$.

To tackle this, we must use the power of calculus. We imagine an infinitesimally thin spherical shell of radius $r$ and thickness $dr$ somewhere inside the material. For this tiny shell, the area is practically constant, so its elemental thermal resistance is:

$$dR = \frac{dr}{K \cdot A}$$

Substituting the surface area of a sphere, $A = \pi r^2$, we get:

$$dR = \frac{dr}{4\pi K r^2}$$

### Integrating the Elemental Resistance

To find the total thermal resistance $R$ of the entire material, we must sum up the resistances of all these infinitesimally thin shells stacked from the inner radius $r_1$ to the outer radius $r_2$. This is exactly what integration does:

$$R = \int_{r_1}^{r_2} \frac{dr}{4\pi K r^2}$$

Pulling the constants out of the integral, we are left with integrating $r^{-2}$ with respect to $r$. The integral of $r^{-2}$ is simply $-\frac{1}{r}$.

$$R = \frac{1}{4\pi K} \left[ -\frac{1}{r} ight]_{r_1}^{r_2}$$

Now, we carefully plug in our upper and lower limits:

$$R = \frac{1}{4\pi K} \left( -\frac{1}{r_2} - \left(-\frac{1}{r_1}ight) ight)$$

$$R = \frac{1}{4\pi K} \left( \frac{1}{r_1} - \frac{1}{r_2} ight)$$

Taking a common denominator, we arrive at the final expression for the total thermal resistance:

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

### The Final Heat Current

With the total thermal resistance in hand, finding the rate of heat flow (or heat current, $\frac{dQ}{dt}$) is straightforward. It is analogous to Ohm's law in electricity ($I = \frac{\Delta V}{R}$). Here, the driving force is the temperature difference $\Delta 	heta = 	heta_2 - 	heta_1$.

$$\frac{dQ}{dt} = \frac{\Delta 	heta}{R}$$

Substituting our expression for $R$:

$$\frac{dQ}{dt} = \frac{	heta_2 - 	heta_1}{\frac{r_2 - r_1}{4\pi K r_1 r_2}}$$

Flipping the denominator to the numerator, we get our final, elegant result:

**$$\frac{dQ}{dt} = \frac{4\pi K r_1 r_2 (	heta_2 - 	heta_1)}{r_2 - r_1}$$**

This formula beautifully captures how the geometry of the spheres dictates the flow of thermal energy.

Select Answer:

Visualized Solution

Visualizing the Setup

Choosing an Elemental Shell

Thermal Resistance of Element

Substituting Surface Area

Integrating for Total Resistance

Evaluating the Integral

Applying Limits

Rate of Heat Flow

Final Answer

The Way Forward

The Sigma Insight: Heat Transfer

The Geometry of the Problem

The Concept of Thermal Resistance

Integrating the Elemental Resistance

The Final Heat Current

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