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Chapter 18: Problem 61

The human body transports heat from the interior tissues, at temperature \(37.0^{\circ} \mathrm{C},\) to the skin surface, at temperature \(27.0^{\circ} \mathrm{C},\) at a rate of \(100 . \mathrm{W}\). If the skin area is \(1.5 \mathrm{~m}^{2}\) and its thickness is \(3.0 \mathrm{~mm}\), what is the effective thermal conductivity, \(\kappa,\) of skin?

Short Answer

Expert verified
Answer: The effective thermal conductivity of the skin is 0.02 W/m·K.

Step by step solution

01

Identify the given values

We are given: - The temperature of the interior tissues, \(T_1 = 37.0 ^{\circ} \mathrm{C}\) - The temperature of the skin surface, \(T_2 = 27.0 ^{\circ} \mathrm{C}\) - The rate of heat transfer, \(Q = 100 \mathrm{W}\) - The skin area, \(A = 1.5 \mathrm{~m}^{2}\) - The thickness of the skin, \(d = 3.0 \mathrm{~mm}\)
02

Convert all values to SI units

To work with the SI units, the temperatures need to be in Kelvin, and the thickness of the skin needs to be in meters. - Convert \(T_1\) and \(T_2\) to Kelvin: \(T_1 = 37.0 ^{\circ} \mathrm{C} + 273.15 = 310.15 \mathrm{K}\) and \(T_2 = 27.0 ^{\circ} \mathrm{C} + 273.15 = 300.15 \mathrm{K}\) - Convert the thickness of the skin to meters: \(d = 3.0 \mathrm{~mm} \times \frac{1 \mathrm{~m}}{1000 \mathrm{~mm}} = 0.003 \mathrm{~m}\)
03

Calculate the temperature difference

Calculate the temperature difference between the interior tissues and the skin surface. \(\Delta T = T_1 - T_2 = 310.15 \mathrm{K} - 300.15 \mathrm{K} = 10 \mathrm{K}\)
04

Rearrange Fourier's Law to find the thermal conductivity

Rearrange Fourier's Law to solve for \(\kappa\): \(\kappa = \frac{-Q \times d}{A \times \Delta T}\)
05

Calculate the effective thermal conductivity

Substitute the given values into the formula to find the effective thermal conductivity of the skin: \(\kappa = \frac{-(-100 \mathrm{W}) \times 0.003 \mathrm{~m}}{1.5 \mathrm{~m}^{2} \times 10 \mathrm{K}} = \frac{0.3 \mathrm{W}}{15 \mathrm{m}^2 \mathrm{K}} = 0.02 \mathrm{W/m} \cdot \mathrm{K}\) The effective thermal conductivity of the skin is \(\kappa = 0.02 \mathrm{W/m} \cdot \mathrm{K}\).

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