a.A person with skin area of 3 m2 is 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)
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~ :)

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

http://en.wikipedia.org/wiki/Natural_convection#Natural_Convection_from_a_Vertical_Plate

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.

To calculate the rate of conduction of heat loss in watts, we need to use the equation for heat conduction:

Q = (k * A * ΔT) / d

Where:
- Q is the rate of heat conduction (in watts)
- k is the thermal conductivity of the material (in watts per meter-kelvin)
- A is the surface area through which the heat is being conducted (in square meters)
- ΔT is the temperature difference between the two ends (in kelvin)
- d is the distance through which the heat is being conducted (in meters)

Since the person is naked, we can assume that the thermal conductivity of air is 0.024 watts per meter-kelvin. The surface area of the person's skin is given as 3 m2, and the temperature difference is 28°C - 22°C = 6°C. The distance through which the heat is being conducted is 5 cm, which is 0.05 meters.

Plugging these values into the equation, we get:

Q = (0.024 * 3 * 6) / 0.05

Simplifying:

Q = 4.32 watts

Therefore, the rate of conduction of heat loss is 4.32 watts.