A child has a temperature of 101◦F.

Question

If her total skin area is 13 m2, find the en-
ergy loss per second due to radiation, assum-
ing the emissivity is 1. The Stefan-Boltzmann
constant is 5.6696 × 10−8 W/m2
· K4 . (As-
sume the room temperature is 70◦F.)
Answer in units of W.

Answer 1

T1= 101◦F =311.48K
T2= 70°F=294.26 K
Since the emissivity is 1, the object is the black body =>
P=R•A= σ•T⁴•A
ΔP= σ•T1⁴•A - σ•T2⁴•A = σ•A•( T1⁴ - T2⁴)=
=5.67•10⁻⁸•13• (311.48⁴ -294.26⁴)=1410 W

Answer 2

شمس