Question

If you lie on the ground at night with no cover, you get cold rather quickly. Much of this is due to energy loss by radiation. At night in a dry climate, the temperature of the sky can drop to −40∘C. If you are lying on the ground with thin clothing that provides little insulation, the surface temperature of your skin and clothes will be about 30∘C. Estimate the net rate at which you body loses energy by radiation to the night sky under these conditions. Hint: What area should we use?

Answer 1

Answer:

301.48 J/s

Explanation:

We are given;

Temperature of the sky dropping to −40∘C: T_o = -40°C = -40 + 273 = 233 K

Temperature of your skin and clothes: T = 30°C = 30 + 273 = 303 K

Body surface area of human body is around 2 m². But here only half of the body is facing the sky, Thus Area is: A = 2/2 = 1 m²

To solve this, we will use the equation for thermal heat transfer known as the Stefan bolt Mann equation.

ΔQ/Δt = εσA(T⁴ - (T_o)⁴)

Where;

ΔQ/Δt is the rate at which you body loses energy by radiation

ε is the emissivity of the human body with a value of 0.97

σ is Stefan boltzmann constant with a value of 5.67 X 10^(-8) W/m².K⁴

Thus;

ΔQ/Δt = 0.97 × 5.67 X 10^(-8) × 1(303⁴ - 233⁴)

ΔQ/Δt = 301.48 J/s