The net rate of radiant heat flow from and object is given by (deltaQ/deltaT) =

Question

The net rate of radiant heat flow from and object is given by (deltaQ/deltaT) = e(sigma)A(T1^4-T2^4)

A: area of object

T1: temperature of object

e: emmisivity (value depends on surface)

T2: temperature of it's surroundings

sigma: 5.67x10^-8

A person has a skin temperature of 33 degrees Celsius and is in a room when the walls have a temperature of 29 degrees Celsius. If the emmisivity for the skin is 0.70 (e=0.70) and if the body surface area is 1.5m^2, then calculate the(deltaQ/deltaT) net, the net radiative heat loss in Watts.

Explanation / Answer

= 5.76*10^(-8)*.70*1.5*(306^4-302^4)

=26.76

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