A metal is heated in a furnace where a sensor is kept above the metal surface to read the power radi

Q: Why can't I just use the temperatures in Celsius?

Stefan's Law is derived from fundamental thermodynamic principles that rely on the absolute zero of temperature. Using Celsius would imply zero radiated power at $0^\circ\text{C}$, which is physically incorrect since objects still radiate heat at freezing temperatures!

Q: What happened to the area and emissivity in Stefan's Law?

The problem states that the emissivity remains constant, and since it's the same metal surface, the area is also constant. When we take the ratio of the powers ($P_2/P_1$), these constant terms ($\sigma$, $e$, $A$) perfectly cancel out.

Q: How did we get $P_1/P_0 = 2$?

The sensor displays $\log_2(P/P_0)$. Initially, this reading is $1$. So, $\log_2(P_1/P_0) = 1$. By the definition of logarithms, if $\log_b(x) = y$, then $x = b^y$. Here, $P_1/P_0 = 2^1 = 2$.

Q: Why did we calculate $P_2/P_1$ instead of finding $P_2$ directly?

We don't have the exact values for area, emissivity, or the constant $P_0$. By finding the ratio $P_2/P_1$, we bypass the need for these unknown constants and can relate the final state directly to the initial state.

Q: Why do engineers use logarithmic sensors?

Radiated power scales with the fourth power of temperature, meaning it grows incredibly fast. A logarithmic scale compresses this massive range of power values into small, manageable numbers (like $1$ to $9$), making it easier to read and process electronically.

Question

A metal is heated in a furnace where a sensor is kept above the metal surface to read the power radiated ($P$) by the metal. The sensor has a scale that displays $\log_2(P/P_0)$, where $P_0$ is a constant. When the metal surface is at a temperature of $487^\circ	ext{C}$, the sensor shows a value $1$. Assume that the emissivity of the metallic surface remains constant. What is the value displayed by the sensor when the temperature of the metal surface is raised to $2767^\circ	ext{C}$?

Sigma Percentile · Accepted Answer

### The Setup: A Furnace, A Metal, and A Logarithmic Sensor

Imagine you are standing in front of an industrial furnace. Inside, a piece of metal is glowing intensely, radiating heat in all directions. Right above this metal surface, we have a specialized sensor designed to capture the radiated power, which we will call $P$.

However, this isn't your everyday sensor. Instead of displaying the raw power in Watts, it has a built-in logarithmic scale. It displays the value of $\log_2(P/P_0)$, where $P_0$ is just some baseline reference constant. This might seem overly complicated at first glance, but as we will see, it is a brilliant engineering trick to handle massive numbers!

### The Master Equation: Stefan's Law

To solve this problem, we need a bridge that connects the temperature of the metal to the power it radiates. Enter **Stefan's Law**.

Stefan's Law tells us that the power $P$ radiated by a black body (or any body with a constant emissivity) is directly proportional to the fourth power of its absolute temperature $T$. Mathematically, we write this as:

$$P = \sigma e A T^4$$

Because the area $A$, the emissivity $e$, and the Stefan-Boltzmann constant $\sigma$ remain unchanged throughout our experiment, we can simplify our lives by just focusing on the core relationship:

$$P \propto T^4$$

**Crucial Trap Warning:** The temperature $T$ in this formula *must* be the absolute temperature in Kelvin. If you plug in Celsius, your entire calculation will collapse!

### The Temperature Transformation

Let's carefully convert our given temperatures from Celsius to Kelvin.

Initially, the metal is at $487^\circ	ext{C}$.

$$T_1 = 487 + 273 = 760 	ext{ K}$$

Later, the furnace is cranked up, and the metal reaches a scorching $2767^\circ	ext{C}$.

$$T_2 = 2767 + 273 = 3040 	ext{ K}$$

### The Power Ratio

Now, let's see how much the radiated power has increased. Since $P \propto T^4$, the ratio of the final power $P_2$ to the initial power $P_1$ is simply the ratio of their absolute temperatures raised to the fourth power:

$$\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}ight)^4$$

Let's substitute our Kelvin temperatures:

$$\frac{P_2}{P_1} = \left(\frac{3040}{760}ight)^4$$

Notice how beautifully the numbers cancel out! $3040$ divided by $760$ is exactly $4$.

$$\frac{P_2}{P_1} = 4^4 = 256$$

This means the metal is now radiating $256$ times more power than it was initially!

### Decoding the Sensor's Logic

We know the ratio of the powers, but we need to figure out what the sensor will actually display. Let's look at the initial condition. We are told that at $T_1$, the sensor reads $1$.

$$\log_2\left(\frac{P_1}{P_0}ight) = 1$$

By converting this logarithmic equation into its exponential form, we can isolate the ratio of $P_1$ to the constant $P_0$:

$$\frac{P_1}{P_0} = 2^1 = 2$$

### The Final Calculation

We are hunting for the final sensor reading, which is $\log_2(P_2/P_0)$. To find this, we need the ratio $P_2/P_0$. We can cleverly construct this by multiplying the two ratios we've already found:

$$\frac{P_2}{P_0} = \frac{P_2}{P_1} 	imes \frac{P_1}{P_0}$$

Substitute the values we calculated:

$$\frac{P_2}{P_0} = 4^4 	imes 2$$

To make the logarithm easy to evaluate, let's express everything in base $2$. Since $4 = 2^2$, we have $4^4 = (2^2)^4 = 2^8$.

$$\frac{P_2}{P_0} = 2^8 	imes 2^1 = 2^9$$

Finally, we plug this back into the sensor's display formula:

$$	ext{Final Reading} = \log_2(2^9)$$

Using the power rule of logarithms, the $9$ drops down to the front, and since $\log_2(2) = 1$, we are left with our final answer:

$$	ext{Final Reading} = 9$$

Isn't it fascinating? The temperature increased by a factor of $4$, the radiated power exploded by a factor of $256$, but thanks to the logarithmic scale, the sensor reading only gently climbed from $1$ to $9$. This is exactly why engineers love logarithmic scales!

Enter Numerical Value:

Visualized Solution

$P_{sensor} = lo g_{2} (\frac{P}{P _{0}})$

$P = σ e A T^{4}$

$T (K) = T (^{\circ} C) + 273$

$\frac{P _{2}}{P _{1}} = (\frac{T _{2}}{T _{1}})^{4}$

$lo g_{2} (\frac{P _{1}}{P _{0}}) = 1$

$\frac{P _{2}}{P _{0}} = \frac{P _{2}}{P _{1}} \times \frac{P _{1}}{P _{0}}$

$Final Reading = lo g_{2} (\frac{P _{2}}{P _{0}})$

\text{Logarithmic Scaling}

The Sigma Insight: Heat Transfer

The Setup

The Master Equation

The Temperature Transformation

The Power Ratio

Decoding the Sensor's Logic

The Final Calculation

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