# SCIENCE

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a.A person with skin area of 3 m2 is nude in a room of still air at 22 C.
i.Calculate the rate of conduction of heat loss in watts if the skin temperature is 28 C at a distance of 5 cm.

• SCIENCE -

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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