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

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