Question

Two surfaces, one highly polished and the other heavily oxidized, are found to be emitting the same amount of energy per unit area. The highly polished surface has an emissivity of \(0.1\) at \(1070^{\circ} \mathrm{C}\), while the emissivity of the heavily oxidized surface is \(0.78\). Determine the temperature of the heavily oxidized surface.

Short answer

Answer: The heavily oxidized surface has a temperature of approximately \(357.29 ^{\circ} \mathrm{C}\).

Step-by-step solution

Understand the Stefan-Boltzmann Law

The Stefan-Boltzmann law states that the energy emitted by a black body per unit area is directly proportional to the fourth power of its temperature (in Kelvin). The equation for Stefan-Boltzmann law is: \(E = e * \sigma * T^4\) where \(E\) represents the energy emitted per unit area, \(e\) is the emissivity of the surface, \(\sigma\) is the Stefan-Boltzmann constant \((5.67 * 10^{-8} \ W / m^2K^4)\), \(T\) is the temperature of the surface in Kelvin. As both surfaces are emitting the same amount of energy per unit area, we can equate their energy emissions and solve for the unknown temperature.

Convert given temperature to Kelvin

The polished surface has a temperature of \(1070^{\circ} \mathrm{C}\). To convert it to Kelvin, we add \(273.15\). \(T_{polished} = 1070 + 273.15 = 1343.15 \ K\)

Set up the equation with given values

Using the Stefan-Boltzmann law, we can set up the equation with the given emissivity values and the known temperature. Let \(T_{oxidized}\) be the temperature of the heavily oxidized surface in Kelvin. \(0.1 \times \sigma \times (1343.15)^4 = 0.78 \times \sigma \times (T_{oxidized})^4\) To simplify, we can divide both sides by \(0.1 \times \sigma\): \((1343.15)^4 = 7.8 \times (T_{oxidized})^4\)

Solve for unknown temperature

To find the value of \(T_{oxidized}\), follow these steps: take the fourth root of both sides, divide by the square root of \(7.8\), and round to the nearest whole number. \((T_{oxidized})^4 = \frac{(1343.15)^4}{7.8}\) \(T_{oxidized} = \sqrt[4]{\frac{(1343.15)^4}{7.8}}\) \(T_{oxidized} \approx 630.44 \ K\)

Convert the temperature to Celsius

To obtain the temperature of the heavily oxidized surface in Celsius, subtract \(273.15\). \(T_{oxidized} = 630.44 - 273.15 = 357.29 ^{\circ} \mathrm{C}\) The heavily oxidized surface has a temperature of approximately \(357.29 ^{\circ} \mathrm{C}\).

Key concepts

Thermodynamics

The science of thermodynamics deals with heat and temperature and their relation to energy and work. It lays out a series of laws that describe how thermal energy is transferred and how it can perform work. This field of physics plays a crucial role when analyzing the energy emitted by different surfaces, such as in the Stefan-Boltzmann Law problem.

Understanding thermodynamics is essential because it dictates that all systems tend towards equilibrium - a state where temperatures are balanced. The discipline isn't just theoretical; it has practical implications in designing energy-efficient materials and systems. By applying thermodynamic principles, we can deduce how different qualities of surfaces, like their finish or composition, influence their energy emission and temperature maintenance capacities.

Emissivity

Emissivity is a measure of a material's ability to emit infrared energy. It is expressed as a ratio from 0 to 1. An emissivity of 1 means the material is a perfect emitter, like a black body, which absorbs and emits all incident thermal radiation, while an emissivity less than 1 suggests that the material reflects some of the incident thermal radiation. The concept is vital when solving problems involving radiative heat transfer, like comparing the emissive properties of a highly polished surface to a heavily oxidized one.

Materials with low emissivity, such as polished metals, are efficient at reflecting heat, while those with high emissivity, like oxidized surfaces, are better at radiating heat away. This has practical applications in a wide range of industries, such as insulation design, thermal imaging, and even astrophysics, for determining celestial body temperatures.

Black Body Radiation

Black body radiation refers to the theoretical perfect absorber and emitter of thermal radiation. In the Stefan-Boltzmann Law, the term black body is used as a standard to describe an ideal emitter. A black body absorbs all electromagnetic radiation, regardless of frequency or angle of incidence, and emits radiation at a maximal rate for its given temperature.

When a real-life object is compared to a black body, we can use the object's emissivity to understand how well it absorbs and emits energy, which is vital for predicting its thermal behavior. This concept is a cornerstone in thermodynamics and helps scientists predict the thermal radiation emitted from surfaces such as the sun, stars, and planets, as well as engineered surfaces like solar cells and thermal insulators.

Temperature Conversion

Temperature conversion between different scales is commonplace in thermodynamics. The Stefan-Boltzmann Law requires temperatures to be in Kelvin, which is the SI base unit for thermodynamic temperature. Converting temperatures accurately from Celsius to Kelvin, as done in the provided exercise, is a fundamental step without which we could not properly use the law or communicate findings.

To convert Celsius to Kelvin, one simply adds 273.15 to the Celsius value. Conversely, Kelvin can be converted back to Celsius by subtracting 273.15. This exercise shows that understanding temperature conversion is not just a mathematical exercise; it's crucial to ensuring the accuracy of thermal calculations and in a broader sense, all scientific communication.