Saturday

January 31, 2015

January 31, 2015

Posted by **tony** on Sunday, October 19, 2008 at 1:00pm.

i just wanted to know if my answer of 74000 j was correct.

- physics -
**drwls**, Sunday, October 19, 2008 at 1:09pmThe 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 Stefan-Boltzmann 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...