Question

3. Body heat is brought by the blood to the bottom layer of skin at a temperature of 37°C, the outer surface of the skin is at 30°C. A 165 lb (75 kg), 6 ft (1.83 m) person has a surface area of about 2 m² and loses heat

at a net rate of 75W while resting. Assuming an average skin thickness of 0.75 mm, what is the thermal conductivity of this person's skin?

A 5x102 W/(m C)
B 4x10 W/(m C")
C 1 x 105 W/(m C")
D 6x 104 W/(m-C)

Answer 1

D

Step-by-step explanation:

To find the thermal conductivity of the person's skin, you can use the formula for heat transfer through a material:

\[Q = \frac{{k \cdot A \cdot \Delta T}}{{d}}\]

Where:

- \(Q\) is the heat transfer rate (75W in this case).

- \(k\) is the thermal conductivity of the material (which we're trying to find).

- \(A\) is the surface area (2 m²).

- \(\Delta T\) is the temperature difference (37°C - 30°C = 7°C).

- \(d\) is the thickness of the material (0.75 mm = 0.00075 m).

Now, plug in the known values and solve for \(k\):

\[75 = \frac{{k \cdot 2 \cdot 7}}{{0.00075}}\]

First, calculate the right side of the equation:

\[k \cdot 2 \cdot 7 = 14k\]

Now, divide both sides by 14:

\[k = \frac{{75}}{{14 \cdot 7/0.00075}}\]

Now, calculate the value:

\[k ≈ 6 \times 10^4 \, \text{W/(m·°C)}\]

So, the thermal conductivity of this person's skin is approximately 6 x 10^4 W/(m·°C), which is option D.