Question

The Sun at the beach or: why do we get sunburns? The Sun impingnes on the ground at 1300 W/m² (time averaged power density). If the properties of skin are known as σ = 0.01 S/m, μ = μ。, and ε = 24ε。, calculate the amount of power dissipated in the skin of a person. Assume the area exposed is 1 m² and the Sun radiates at an average frequency of 5 x 10¹⁴ Hz and is perpendicular to the surface of the skin.

Answer 1

The question involves calculating power dissipated in skin due to sun exposure. With an intensity of sunlight at 1300 W/m² and exposed skin area of 1 m², the power absorbed by skin would be 1300 Watts, assuming 100% absorption.

Step-by-step explanation:

The student's question pertains to the physics of sunlight and its interaction with the human skin, specifically focusing on the power dissipated in the skin due to sun exposure. Given the time-averaged power density of the Sun at 1300 W/m², skin properties of σ = 0.01 S/m, μ = μ、, ε = 24ε、, and the area exposed to sunlight is 1 m², we would calculate the power absorbed and subsequently dissipated in the skin.

While the question provides details like the frequency of the Sun's radiation and the permittivity (ε), these values aren't directly needed for calculating the power dissipation in the skin using the provided parameters.

Under typical circumstances, we use the intensity of the sunlight and the area to determine the power absorbed by the skin. With an exposure area of 1 m² and assuming all the incident sunlight is absorbed, the power absorbed would simply be the intensity times the area, which would be 1300 W/m² × 1 m² = 1300 W.

The properties given for skin conductivity (σ) and permittivity (ε) are more relevant for electromagnetic analyses and do not directly apply to the calculation of power dissipation due to sunlight exposure.