One end of a rod of length L and cross-sectional area A is kept in a furnace of temperature T1. The

Q: Why do we ignore heat loss from the sides of the rod?

The problem explicitly states the rod is insulated on the sides and heat is lost *only* by radiation at the end where the temperature is $T_2$.

Q: Why can we use the binomial expansion here?

Because $\Delta T \ll T_s$, the ratio $\frac{\Delta T}{T_s}$ is very small (much less than 1). The binomial expansion $(1+x)^n \approx 1+nx$ is highly accurate for $x \ll 1$.

Q: What happened to the area $A$ in the equation?

The cross-sectional area $A$ appears on both sides of the steady-state heat balance equation (in both the conduction and radiation terms) and thus cancels out.

Q: Is this related to Newton's Law of Cooling?

Yes! Newton's Law of Cooling is essentially the linearized form of the Stefan-Boltzmann law for small temperature differences, which is exactly what we derived here.

Q: What if $\Delta T$ was not small?

If $\Delta T$ were large, we could not use the binomial approximation. The relationship between $\Delta T$ and $(T_1 - T_s)$ would remain non-linear, involving fourth-order polynomial roots.

Question

One end of a rod of length $L$ and cross-sectional area $A$ is kept in a furnace of temperature $T_1$. The other end of the rod is kept at a temperature $T_2$. The thermal conductivity of the material of the rod is $K$ and emissivity of the rod is $e$.
It is given that $T_2 = T_s + \Delta T$, where $\Delta T \ll T_s$, $T_s$ being the temperature of the surroundings. If $\Delta T \propto (T_1 - T_s)$, find the proportionality constant. Consider that heat is lost only by radiation at the end where the temperature of the rod is $T_2$.

Sigma Percentile · Accepted Answer

The problem presents a fascinating interplay between two fundamental modes of heat transfer: **conduction** and **radiation**. We have a rod acting as a thermal bridge between a hot furnace at temperature $T_1$ and the cooler surroundings at temperature $T_s$.

The key to unlocking this problem lies in understanding the steady-state condition and applying a clever mathematical approximation.

### Analyzing the Setup

In a steady state, the temperature profile of the rod does not change with time. This means that the rate at which heat enters the rod from the furnace must equal the rate at which heat travels through the rod, which in turn must equal the rate at which heat is lost from the other end.

The problem explicitly states that heat is lost *only* by radiation at the exposed end (temperature $T_2$). Therefore, we can equate the rate of heat conduction through the rod to the rate of heat radiation from its end.

The rate of heat conduction is given by Fourier's Law:
$$H_{cond} = \frac{KA(T_1 - T_2)}{L}$$

The rate of heat radiation is given by the Stefan-Boltzmann Law:
$$H_{rad} = eA\sigma(T_2^4 - T_s^4)$$

Equating the two, we get our master equation:
$$\frac{KA(T_1 - T_2)}{L} = eA\sigma(T_2^4 - T_s^4)$$

Notice how the cross-sectional area $A$ beautifully cancels out, leaving us with:
$$\frac{K(T_1 - T_2)}{L} = e\sigma(T_2^4 - T_s^4)$$

### The Binomial Magic

We are given that $T_2 = T_s + \Delta T$, where $\Delta T \ll T_s$. This small temperature difference is our cue to use an approximation. Let's substitute $T_2$ into the radiation term:
$$T_2^4 = (T_s + \Delta T)^4$$

To make this manageable, we factor out $T_s$:
$$T_2^4 = T_s^4 \left(1 + \frac{\Delta T}{T_s}\right)^4$$

Since $\frac{\Delta T}{T_s} \ll 1$, we can apply the binomial approximation $(1+x)^n \approx 1+nx$:
$$T_2^4 \approx T_s^4 \left(1 + 4\frac{\Delta T}{T_s}\right) = T_s^4 + 4T_s^3 \Delta T$$

This simplifies the radiation term immensely:
$$T_2^4 - T_s^4 \approx 4T_s^3 \Delta T$$

### Final Calculation

Now, let's substitute this back into our master equation, and also replace $T_2$ with $T_s + \Delta T$ on the conduction side:
$$\frac{K(T_1 - (T_s + \Delta T))}{L} = e\sigma(4T_s^3 \Delta T)$$

Expanding the left side:
$$\frac{K(T_1 - T_s)}{L} - \frac{K\Delta T}{L} = 4e\sigma T_s^3 \Delta T$$

Our goal is to isolate $\Delta T$. Let's move all terms containing $\Delta T$ to one side:
$$\frac{K(T_1 - T_s)}{L} = 4e\sigma T_s^3 \Delta T + \frac{K\Delta T}{L}$$

Factoring out $\Delta T$:
$$\frac{K(T_1 - T_s)}{L} = \left(4e\sigma T_s^3 + \frac{K}{L}\right) \Delta T$$

Taking the common denominator on the right side:
$$\frac{K(T_1 - T_s)}{L} = \left(\frac{4e\sigma L T_s^3 + K}{L}\right) \Delta T$$

Finally, solving for $\Delta T$:
$$\Delta T = \left(\frac{K}{4e\sigma L T_s^3 + K}\right) (T_1 - T_s)$$

The problem states that $\Delta T \propto (T_1 - T_s)$. By comparing our result with this relation, we can clearly see that the proportionality constant is:
**$$\frac{K}{4e\sigma L T_s^3 + K}$$**

This result is profound. It shows how a highly non-linear process like radiation can be linearized for small temperature differences, effectively defining a "radiative thermal resistance" that acts in series with the conductive thermal resistance of the rod.

Visualized Solution

\text{Steady State Heat Balance}

\text{Substituting } T_2

\text{Binomial Approximation}

\text{Updating the Equation}

\text{Solving for } \Delta T

\text{Final Proportionality Constant}

\text{The Way Forward}

The Sigma Insight: Heat Transfer

Analyzing the Setup

The Binomial Magic

Final Calculation

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