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 34.8°C, the thickness of the tissues between averages 1.00 cm, and the surface area is 1.75 m2. The conductivity of tissue is 0.20 J / (s · m · °C).

The answer is 105 W

Explanation:

Solution

Given that:

The core temperature is = 37.0°C

The skin temperature is = 34.8°C,

The thickness of the tissues is on the average of = 1.00 cm

The surface area = 1.75 m²

The conductivity of tissue is = 0.20 J (s. m. °C)

Now,

The rate of heat transfer through a means of conduction of materials is represented as follows:

Q / t = kA (T₂ - T₁) / d

Where

k = the thermal conductivity

A = denoted as the surface area

d = the thickness

(T₂ - T₁) = is the temperature difference of the body

Thus

Q/t = kA (T₂ - T₁) / d

We now substitute the values of 1.75 m² for A, 0.20 J / (s. m. °C) for k, 37.0°C, for T₁, 34.8°C for T₂, 1.00 cm for d

We have the following inputs:

Q/t = (0.20 J/s. m. °C) (1.75 m²) (37.0°C - 34.8°C) / ((1 cm) (10^⁻2/1 cm))

Now

Q/t = (0.20 J/s. m. °C) (1.75 m²) (3.00 ° C) / (0.01 m)

Q/t = 1.05 / (0.01)

Q/t = 105 J/s

Therefore, the rate of heat conduction is 105 W