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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