Posted by tony on Sunday, October 19, 2008 at 1:00pm.
A student is trying to decide what to wear. His bedroom is at 20.0°C. His skin temperature is 36.0°C. The area of his exposed skin is 1.46 m2. People all over the world have skin that is dark in the infrared, with emissivity about 0.915. Find the net energy loss from his body by radiation in 13.5 min.
i just wanted to know if my answer of 74000 j was correct.

physics  drwls, Sunday, October 19, 2008 at 1:09pm
The radiative heat loss RATE of his body is
(Area)*(infrared emissivity)*(sigma)*(T2^4  T1^4)
where T2 = 309 K and T1 = 293 K
(Temperature must be in K to use this formula).
sigma is the StefanBoltzmann constant, which you should have learned about.
There will also be a convective heat loss, but that is harder to calculate. It could be comparable to the radiative loss.
You will have to multiply the heat loss rate by 13.5 min*60 s/min to get the energy loss in that time.
Answer This Question
Related Questions
 physics  A person with skin area 2 m2 and 0.97 radiation efficiency is at rest ...
 physics  A person with skin area 2 m2 and 0.97 radiation efficiency is at rest ...
 physics  A person with skin area 2 m2 and 0.97 radiation efficiency is at rest ...
 physics  An unclothed student is in 24 ◦ C room. If the skin temperature ...
 physics  An unclothed student is in a 27 degree Celsius room. If the skin ...
 physics  a. A person with skin area 2 m2 and 0.97 radiation efficiency is at ...
 physics  a. A person with skin area 2 m2 and 0.97 radiation efficiency is at ...
 PHYSICS  A person with skin area 2 m2 and 0.97 radiation efficiency is at rest ...
 PHYSICS  A person with skin area 2 m2 and 0.97 radiation efficiency is at rest ...
 PHYSICS  a.A person with skin area of 3 m2 is nude in a room of still air at 22...
More Related Questions