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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer
Heat transfer involves the movement of thermal energy from one place to another, and there are three main modes of heat transfer:
  • Conduction: The transfer of heat through a material without the movement of the material itself.
  • Convection: The transfer of heat by the movement of fluid (liquid or gas).
  • Radiation: The transfer of heat in the form of electromagnetic waves.

In the context of the exercise, we focus on heat transfer by conduction through the skin. Conduction occurs when there is a temperature difference between two points. Here, heat moves from the warm interior tissues to the cooler skin surface. Understanding this transfer is crucial as it helps in calculating the thermal properties of materials, like the thermal conductivity of skin.
Fourier's Law
Fourier's Law is a fundamental principle that describes the conduction of heat. It states that the rate of heat transfer through a material is proportional to the negative gradient of temperatures and the thermal conductivity of the material.

Fourier's Law can be mathematically expressed as:\[Q = -abla(T) imes A imes abla d \]Where:
  • \( Q \) is the rate of heat transfer (Watts).
  • \( abla(T) \) is the temperature gradient (difference in temperature over distance).
  • \( A \) is the area through which heat is flowing (square meters).
  • \( abla d \) is the direction of heat flow.

In practical terms, the law helps to connect how efficiently a material can conduct heat, given its intrinsic properties. For the given problem, Fourier’s Law is used to determine the effective thermal conductivity.
SI Units Conversion
Working with SI units is essential in physics to ensure consistent calculations. In the exercise, certain values are given in typical units, like degrees Celsius and millimeters, which need conversion to Kelvin and meters respectively for calculations.
  • Temperature in Celsius can be converted to Kelvin by adding 273.15. For example, \(37.0 ^{\circ} C\) becomes \(310.15 K\).
  • Length in millimeters can be converted to meters by multiplying by the factor \( \frac{1 \text{ m}}{1000 \text{ mm}} \). Hence, \(3.0 \text{ mm}\) turns into \(0.003 \text{ m}\).

Ensuring all your measurements are in the appropriate SI units is crucial for standardizing calculations and communicating results effectively. This conversion is an important step especially in fields like heat transfer, where precision is key.
Temperature Difference
The temperature difference between two points is a driving force for heat transfer. When a heat source, like the human body, has a different temperature from its surroundings, heat flows from the hotter area to the cooler one to achieve equilibrium.

To calculate temperature difference, subtract the temperature at one point from the temperature at another:\[\Delta T = T_1 - T_2\]Where:
  • \( T_1 \) is the temperature at the interior tissues expressed in Kelvin.
  • \( T_2 \) is the temperature at the skin surface also in Kelvin.

In this exercise, the temperature difference was found to be 10 K (Kelvin). This difference is crucial as it determines the rate at which heat is conducted through the skin, impacting the calculation of its thermal conductivity.

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Most popular questions from this chapter

An air-cooled motorcycle engine loses a significant amount of heat through thermal radiation according to the Stefan-Boltzmann equation. Assume that the ambient temperature is \(T_{0}=27^{\circ} \mathrm{C}(300 \mathrm{~K})\). Suppose the engine generates 15 hp \((11 \mathrm{~kW})\) of power and, due to several deep surface fins, has a surface area of \(A=0.50 \mathrm{~m}^{2}\). A shiny engine has an emissivity \(e=0.050\), whereas an engine that is painted black has \(e=0.95 .\) Determine the equilibrium temperatures for the black engine and the shiny engine. (Assume that radiation is the only mode by which heat is dissipated from the engine.)

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A gas enclosed in a cylinder by means of a piston that can move without friction is warmed, and 1000 J of heat enters the gas. Assuming that the volume of the gas is constant, the change in the internal energy of the gas is a) 0 . b) 1000 J. c) -1000 J. d) none of the above.

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Which of the following statements is (are) true? a) When a system does work, its internal energy always decreases. b) Work done on a system always decreases its internal energy. c) When a system does work on its surroundings, the sign of the work is always positive. d) Positive work done on a system is always equal to the system's gain in internal energy. e) If you push on the piston of a gas-filled cylinder, the energy of the gas in the cylinder will increase.

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