Posted by KAIT on Thursday, August 18, 2011 at 4:13am.
recall that the rate of energy being transferred is given by:
q = -kA(dT/dx)
where
q = energy per unit time (in Watts)
k = thermal conductivity (W/m-K)
T = temperature (in Kelvin)
x = distance (in meters)
assuming steady-state heat transfer (meaning, at a particular location, the variable does not vary with time), the formula becomes:
q = -kA(T2 - T1)/(x2 - x1)
where subscripts 2 and 1 refer to final and initial values, respectively.
for air, k = 0.025 W/(m-K)
substituting,
q = -0.025*3*(22-28)/(0.05)
q = 0.0225 W
..this is the amount of heat that is being transferred from the skin to air at a distance of 0.05 meters.
*note that for change in temp, you may not convert Celsius to Kelvin since the conversion factor (which is, + 273) will just cancel each other, and answers will be the same:
in Kelvin: (22+273) - (28+273) = 22-28
hope this helps~ :)
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