Post a New Question


posted by .

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.


    recall that the rate of energy being transferred is given by:
    q = -kA(dT/dx)
    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)
    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~ :)


    I have no idea what the 5 cm "distance" represents. You cannot assume that the heat is conducted through a 5 cm layer of air, with a linear temperature gradient in between. That is not what happens. This is a natural convection problem. There is a dimensionless equation involving a parameter called the "Grashof Number" that will allow you to estimate the natural convection heat loss. It should be find a reference to it online.


    For a discussion of how to perform a natural convection problem correctly, see

    What you want to calculate, to estimate the heat transfer, is a Newtonian "film coefficient" h or a "Nusselt number" Nu. To do this will require calculating the Grashof number first.

    You may get about the same answer as you would get assuming heat is concucted through a 5 cm layer of still air, but this would be a lucky coincidence.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question