Problem 58 - College Physics

Question

A box-shaped coal-burning stove has exhausted most of its fuel, and its surface temperature has fallen to \(27^{\circ} \mathrm{C}\). After more coal is added, the surface temperature eventually rises to \(327^{\circ} \mathrm{C}\). By what factor does the stove's radiation heat transfer to the surroundings increase after the coal is added?

Step-by-Step Solution

Answer

The radiation heat transfer increases by a factor of approximately 16.

Step 1: Understanding the Problem

We need to find the factor by which the stove's radiation heat transfer increases following the addition of more coal. This involves understanding how radiation heat transfer works and is affected by temperature changes.

Step 2: Identify the Relevant Formula

Radiation heat transfer is given by Stefan-Boltzmann law: \[ Q = \varepsilon \sigma A T^4 \]where \( Q \) is the radiated power, \( \varepsilon \) is the emissivity, \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the surface area, and \( T \) is the temperature in Kelvin.

Step 3: Convert Temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin since radiation calculations need temperatures in Kelvin.\[ T_1 = 27^{\circ}C = 27 + 273.15 = 300.15 \, K \]\[ T_2 = 327^{\circ}C = 327 + 273.15 = 600.15 \, K \]

Step 4: Calculate Initial and Increased Radiation

Using the formula for radiation power, calculate the radiation at both temperatures:\[ Q_1 = \varepsilon \sigma A (300.15)^4 \] \[ Q_2 = \varepsilon \sigma A (600.15)^4 \]

Step 5: Determine the Factor Increase

The factor by which radiation heat transfer increases is the ratio of \( Q_2 \) to \( Q_1 \):\[ \text{Factor} = \frac{Q_2}{Q_1} = \frac{(600.15)^4}{(300.15)^4} \]Simplifying this, we get:\[ \text{Factor} = \left( \frac{600.15}{300.15} \right)^4 \approx 16 \]

Step 6: Finalize the Answer

The factor by which the radiation heat transfer increases, when the surface temperature of the stove rises from \(27^{\circ}C\) to \(327^{\circ}C\), is approximately 16.

Key Concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a fundamental principle in physics that relates to radiation heat transfer. It states that the total energy emitted per unit surface area of a black body in unit time is directly proportional to the fourth power of the black body's absolute temperature. This relationship is crucial in calculating how much heat energy an object radiates based on its temperature.
This law is mathematically expressed as follows:

• \( Q = \varepsilon \sigma A T^4 \)

Where:

• \( Q \) represents the radiated power, or energy emitted per unit time.
• \( \varepsilon \) is the emissivity of the object, a measure of how efficiently it emits radiation compared to a perfect black body.
• \( \sigma \) is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4\).
• \( A \) is the surface area of the object, which indicates how big the radiating surface is.
• \( T \) is the absolute temperature in Kelvin.

Understanding this law helps us determine the changes in radiative power when there's a temperature shift, as seen in the original exercise involving a coal-burning stove.

Temperature Conversion to Kelvin

When dealing with radiation heat transfer, it is essential to use the absolute temperature scale, which is Kelvin. The Kelvin scale is an absolute measure of temperature starting from absolute zero, the point where no thermal energy remains in a substance. Unlike the Celsius scale, Kelvin avoids negative temperatures, making calculations straightforward.
To convert from degrees Celsius to Kelvin, simply add \(273.15\) to the Celsius temperature. This conversion is vital for directly applying the Stefan-Boltzmann law, as it requires temperature in Kelvin.
In the exercise, the stove's temperatures were given:

• Initial Temperature: \(27^{\circ} \text{C} \to 27 + 273.15 = 300.15 \, \text{K}\)
• Final Temperature: \(327^{\circ} \text{C} \to 327 + 273.15 = 600.15 \, \text{K}\)

Appropriate temperature conversion ensures accurate calculation of radiative power changes.

Emissivity

Emissivity is a material property that measures how effectively a surface emits thermal radiation compared to an ideal black body. A perfect black body, which is a theoretical object that absorbs all incident radiation, has an emissivity of 1. Most real-world objects have emissivity values between 0 and 1.

• \( \varepsilon = 1 \): Perfect emitter (black body)
• \( \varepsilon = 0 \): No emission (perfect reflector)
• \( 0 < \varepsilon < 1 \): Real material

In the context of the radiation heat transfer exercise, emissivity (\(\varepsilon\)) is part of the Stefan-Boltzmann law equation and impacts the calculation of radiated power (Q). Knowing the emissivity of the stove material allows us to understand how much of the theoretical maximum radiation is actually emitted. Although the exercise does not provide an explicit value for emissivity, it is implied for understanding the general principle of how radiative power is calculated.

Radiated Power Calculation

Radiated power calculation involves determining the amount of thermal energy emitted by an object as radiation, relying on the Stefan-Boltzmann law. This calculation depends on several factors: the object's emissivity, surface area, and absolute temperature in Kelvin.
In practice, to find the change in radiation power due to temperature changes, follow these steps:

• Convert all temperatures to Kelvin, ensuring consistency with the formula.
• Calculate the radiated power at each temperature using \( Q = \varepsilon \sigma A T^4 \).
• Determine the factor increase by dividing the radiated power at the higher temperature by that at the lower temperature.

Using the provided exercise, the initial and final temperatures were 300.15 K and 600.15 K, respectively. The radiated power increased by a factor of approximately 16 after more coal was added, demonstrating the strong dependence of radiation on temperature. This significant increase highlights the importance of the temperature variable in radiation calculations.