Question

5) A student is trying to decide what to wear. His bedroom is at 20.0°C. His skin temperature is 35.0°C. The area of his exposed skin is 1.50 m2 . People all over the world have skin that is dark in the infrared, with emissivity about 0.900. Find the net energy transfer from his body by radiation in 10.0 min. (4%)

Answer 1

Answer:

1247.24J or 1.25kJ

Explanation:

The rate of heat transfer by emitted radiation is determined by the Stefan-Boltzmann law of radiation:

Qnet/t =&eA(Ts⁴-Ta⁴)

Qnet/t is the net energy transfer per time

& is the Stefan-Boltzmann constant 5.671 × 10^-8 kg s^-3 K-4 (try to use correct sign)

e is the emissivity of the object i.e is a measure of how well it radiates. In this case, 0.9

A is the area of the radiating surface; 1.5m²

Ts is the absolute skin temperature: 35+273=308K

Ta is the absolute ambient (bedroom) temperature: 20+273=293K

Qnet/t= 5.671×10^-8 ×0.9×1.5×(308⁴-293⁴)

Qnet/t=124.724J/s

For 10 mins or 600sec

Qnet=124.724×t

Qnet= 1247.24J or 1.25kJ