Question

Calculate the rate of heat conduction out of the human body, assuming that the core internal temperature is 37.0°C, the skin temperature is 32.8°C, the thickness of the tissues between averages 1.00 cm, and the surface area is 1.90 m2. The conductivity of tissue is 0.20 J/(s · m · °C).

Answer 1

The rate of heat conduction out of the human body is: 159.6(J/s)

Step-by-step explanation:

We need to apply the thermal conduction equation (
`(Q)/(t) =(K*A*(T_(2)-T_(1)) )/(L)`), where Q is heat, t is time, K is the thermal conductivity constant of the material, A is the surface area of transfer heat, T is temperature and L is the distance of the "wall" between the differents temperatures (see attached), it's used because the tissues are in direct contact between them. Now we need to know that Q/t is the rate of heat conduction so replacing the values given, we get:
`(Q)/(t)=(0.2*1.9*(32.8-37))/(0.01) =-159.6(J/s)`, the negative sign means the heat is outing of the body, so the rate of heat conduction out of the human body will be: 159.6(J/s).