Question

The temperature of your skin is approximately \(35.0^{\circ} \mathrm{C}\). a) Assuming that your skin is a blackbody, what is the peak wavelength of the radiation it emits? b) Assuming a total surface area of \(2.00 \mathrm{~m}^{2}\), what is the total power emitted by your skin? c) Given your answer to part (b), why don't you glow as brightly as a light bulb?

Short answer

Answer: The peak wavelength of the radiation emitted by the skin is 9.41 μm, and the total power emitted is 1001.5 W. The skin doesn't glow as brightly as a light bulb because it emits most of its radiation in infrared wavelengths, which are not visible to the human eye.

Step-by-step solution

(Step 1: Calculate Peak Wavelength using Wien's Law)

First, we need to determine the peak wavelength emitted by the skin using Wien's Displacement Law. Wien's Law states that \(\lambda_{max} = \frac{b}{T}\), where \(b\) is Wien's constant \((2.90 \times 10^{-3} \, \mathrm{m\cdot K})\) and \(T\) is the temperature in Kelvin. First, convert the skin temperature from Celsius to Kelvin: \(T = 35.0 + 273.15 = 308.15 \, \mathrm{K}\). Now, apply Wien's Law to find the peak wavelength: \(\lambda_{max} = \frac{2.90 \times 10^{-3}}{308.15} = 9.41 \times 10^{-6} \, \mathrm{m}\) = 9.41 \, \mathrm{\mu m}$

(Step 2: Calculate Total Power using Stefan-Boltzmann Law)

Next, we need to determine the total power emitted by the skin using Stefan-Boltzmann Law. The law states that \(P = Ae\sigma T^4\), where \(A\) is the surface area, \(e\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \, \mathrm{W\cdot m^{-2}\cdot K^{-4}})\) and \(T\) is the temperature in Kelvin. We are given the surface area \(A = 2.00\, \mathrm{m^2}\) and can assume the emissivity of skin as \(e = 1\) (as it's a blackbody). Now, apply Stefan-Boltzmann Law to find the total power: \(P = (2.00) \times 1 \times (5.67 \times 10^{-8}) \times (308.15)^4 = 1001.5 \, \mathrm{W}\).

(Step 3: Explain why skin doesn't glow as brightly as a light bulb)

Despite the calculated power of 1001.5 Watts, the skin doesn't glow as brightly as a light bulb because the peak wavelength of the radiation emitted by the skin is in the infrared range (9.41 \(\mu\)m). This means that the majority of the radiation emitted by the skin is not visible to the human eye. Light bulbs emit most of their radiation in visible wavelengths, making them appear bright. In summary: a) The peak wavelength of the radiation emitted by the skin is 9.41 \(\mu\)m. b) The total power emitted by the skin is 1001.5 W. c) The skin doesn't glow as brightly as a light bulb because it emits most of its radiation in infrared wavelengths which are not visible to the human eye.

Key concepts

Wien's Displacement Law

Wien's Displacement Law provides a fascinating insight into the relationship between the temperature of an object and the wavelength of radiation it emits most intensely. According to this principle, the hotter the object, the shorter the wavelength of its peak emission. The formula to calculate this peak wavelength is quite simple:
\[\begin{equation} \[2pt]\[2pt]lambda_{max} = \frac{b}{T} \[2pt]\[2pt]\end{equation}\]
where \(lambda_{max}\) represents the peak wavelength, \(b\) is the constant value of approximately \(2.90 \times 10^{-3} m \cdot K\), and \(T\) is the absolute temperature measured in Kelvin. When examining human skin, which has a temperature around 308.15 K, the peak emission occurs in the infrared region. This is pivotal to understanding why our skin does not emit visible light, even though it indeed radiates energy.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law allows us to calculate the total energy emitted per unit surface area of a blackbody. The law can be expressed with the equation:
\[\begin{equation} \[2pt]\[2pt]P = Ae\sigma T^4 \[2pt]\[2pt]\end{equation}\]
where \(P\) is the total emitted power, \(A\) is the surface area, \(e\) represents the emissivity of the material, \(\sigma\) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8} W\cdot m^{-2}\cdot K^{-4}\), and \(T\) is the absolute temperature in Kelvin. For an ideal blackbody, such as our skin being approximated as one, the emissivity (\(e\)) is 1. This simplifies the equation and helps us determine how much energy is emitted based on the surface area and temperature. For human skin, the power emitted is surprisingly high, but again, understanding emissivity is key to knowing why we do not illuminate like a traditional light source.

Infrared Radiation

Infrared radiation is a type of electromagnetic radiation with wavelengths longer than visible light but shorter than microwaves. This invisible form of energy is what human skin predominantly emits due to its temperature. Objects at room temperature also emit infrared radiation, which can be detected by infrared cameras and sensors.
Infrared is divided into subcategories: near, mid, and far-infrared. The nearer to the visible spectrum, the shorter the infrared wavelength. Since the peak wavelength of human skin's radiation falls within the far-infrared range, around 9.41 micrometers, it is far from the spectrum of visible light, explaining why our skin does not visibly glow despite emitting energy discernible by certain instruments or creatures, such as snakes that can sense infrared.

Emissivity

Emissivity is a measure of how efficiently an object emits thermal radiation compared to a perfect blackbody. It is a dimensionless factor ranging from 0 to 1, with 1 indicating an ideal blackbody. Most natural objects have emissivities less than 1, reflecting their imperfection in radiating energy.
An object's emissivity affects its observed temperature when measured with thermal imaging cameras. For instance, shiny metal surfaces can have a low emissivity and thus appear cooler than they actually are. Since human skin is approximated as a blackbody, we use an emissivity of 1 when applying the Stefan-Boltzmann Law in calculations. However, in real-world applications, the emissivity would be slightly less, impacting the total power calculations.