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~ :)

To calculate the rate of conduction of heat loss, we can use the formula:

Q/t = k * A * ΔT / d

Where:
Q/t = rate of heat conduction
k = thermal conductivity of air
A = area of skin
ΔT = temperature difference
d = distance

First, we need to find the thermal conductivity of air at 22°C. You can either refer to a thermal conductivity table or use an approximate value of 0.024 W/m∙K.

Next, we need to calculate the temperature difference (ΔT).

ΔT = T_final - T_initial
= 28°C - 22°C
= 6°C

Now we can substitute the given values into our formula:

Q/t = k * A * ΔT / d

Q/t = 0.024 W/m∙K * 3 m² * 6°C / 0.05 m

Q/t = 0.024 * 3 * 6 / 0.05
= 8.64 W

So, the rate of conduction of heat loss in watts is approximately 8.64 W.