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Consider a person standing in a room at \(20^{\circ} \mathrm{C}\) with an exposed surface area of \(1.7 \mathrm{~m}^{2}\). The deep body temperature of the human body is \(37^{\circ} \mathrm{C}\), and the thermal conductivity of the human tissue near the skin is about \(0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The body is losing heat at a rate of \(150 \mathrm{~W}\) by natural convection and radiation to the surroundings. Taking the body temperature \(0.5 \mathrm{~cm}\) beneath the skin to be \(37^{\circ} \mathrm{C}\), determine the skin temperature of the person.

Short Answer

Expert verified
Answer: The skin temperature of the person is approximately \(31.25^{\circ}\mathrm{C}\).

Step by step solution

01

Write down the known values

Deep body temperature (\(T_1\)): \( 37^{\circ}\mathrm{C}\) Temperature beneath the skin (\(T_1\)): \(37^{\circ}\mathrm{C}\) Temperature of the room (\(T_2\)): \(20^{\circ}\mathrm{C}\) Exposed surface area (\(A\)): \(1.7\mathrm{m}^2\) Thermal conductivity (\(k\)): \(0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) Heat loss rate (\(Q\)): \(150\mathrm{W}\) Distance beneath the skin (\(d\)): \(0.5\mathrm{cm}\) or \(0.005\mathrm{m}\)
02

Convert temperature values to Kelvin

To ensure all values are consistent and since the heat transfer equation uses Kelvin, we need to convert the temperatures to Kelvin. \(T_1 (K) = 37^{\circ}\mathrm{C} + 273.15 = 310.15\mathrm{K}\) \(T_2 (K) = 20^{\circ}\mathrm{C} + 273.15 = 293.15\mathrm{K}\)
03

Set up the equation for heat transfer by conduction

The heat transfer by conduction can be calculated using Fourier's Law: \(Q = kA \frac{T_1 - T_{skin}}{d}\) Where \(Q\) is the heat loss rate, \(k\) is the thermal conductivity, \(A\) is the exposed surface area, \(T_1\) is the temperature beneath the skin, \(T_{skin}\) is the skin temperature, and \(d\) is the distance beneath the skin.
04

Solve for the skin temperature (\(T_{skin}\))

We want to find \(T_{skin}\), so we need to rearrange the equation from step 3: \(T_{skin} = T_1 - \frac{Qd}{kA}\) Now plug in the given values: \(T_{skin} = 310.15\mathrm{K} - \frac{150\mathrm{W} \cdot 0.005\mathrm{m}}{0.3 \mathrm{W} / \mathrm{m} \cdot \mathrm{K} \cdot 1.7\mathrm{m}^2}\)
05

Calculate the skin temperature and convert back to Celsius

Perform the calculation: \(T_{skin} = 310.15\mathrm{K} - \frac{150\mathrm{W} \cdot 0.005\mathrm{m}}{0.3 \mathrm{W} / \mathrm{m} \cdot \mathrm{K} \cdot 1.7\mathrm{m}^2} = 304.4\mathrm{K}\) Convert back to Celsius: \(T_{skin}^{\circ}\mathrm{C} = 304.4\mathrm{K} - 273.15 = 31.25^{\circ}\mathrm{C}\) The skin temperature of the person is approximately \(31.25^{\circ}\mathrm{C}\).

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Key Concepts

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

Fourier's Law
Fourier's Law is a fundamental principle that describes how heat is propagated through solid materials due to a temperature gradient. It is named after the French mathematician and physicist Joseph Fourier, who formulated it.

In the context of heat transfer in the human body, Fourier's Law can be applied to determine the rate of heat conduction through layers of tissue. This law is formally expressed by the equation: \[\begin{equation}Q = -kA \frac{dT}{dx}\end{equation}\]Where:
  • Q represents the heat flow rate in watts (W).
  • k is the thermal conductivity of the material (W/m·K).
  • A is the cross-sectional area perpendicular to heat flow (m2).
  • dT is the temperature difference between two points (K).
  • dx is the thickness of the material through which heat is flowing (m).
The negative sign indicates that heat flows in the direction of decreasing temperature. In practical terms, Fourier's Law tells us that heat transfer through the human tissue is proportional to the thermal gradient and the tissue's area and thermal conductivity. Understanding Fourier's Law is vital to calculating how quickly the body loses heat in different environmental conditions.
Thermal Conductivity
Thermal conductivity, denoted by the symbol k, is a material-specific property that measures a material's ability to conduct heat. It describes the ease with which heat can pass through a material and is measured in units of watts per meter per Kelvin (W/m·K).

Regarding the human body, the thermal conductivity of human tissue is important because it dictates how readily heat can move from the warmer interior of the body to the cooler exterior environment, or vice versa. In the given exercise, a thermal conductivity value of 0.3 W/m·K for human tissue is specified. This indicates a relatively low rate of thermal energy transfer, which is why humans are capable of maintaining a stable internal temperature despite varying external temperatures.

Higher thermal conductivity means greater efficiency in heat transfer, as seen in metals like copper and aluminum. In contrast, materials like air, fat, and layers of clothing have lower thermal conductivities and thus, serve as insulators, slowing the rate of heat loss from the body.
Convection and Radiation
Convection and radiation are two distinct mechanisms of heat transfer that play significant roles in thermal regulation of the human body. Unlike conduction, which involves transfer of heat through a solid medium, convection and radiation can occur through fluids (gases and liquids) and vacuum, respectively.

Convection is the heat transfer due to the movement of a fluid, such as air or water, over the surface of a body. When the body is hotter than the surrounding air, it will transfer heat to the air. The warmed air then moves away from the skin, replaced by cooler air, which is subsequently heated. This mode of heat transfer is influenced by factors such as the difference in temperature between the body and the environment, as well as the velocity of the moving fluid.

Radiation, on the other hand, is heat transfer that occurs in the form of electromagnetic waves, primarily in the infrared spectrum. Any object with a temperature above absolute zero emits radiation; a human body does this primarily via infrared radiation. The rate of heat loss by radiation is influenced by the temperature of the body, the temperature of the surrounding surfaces, and the emissivity of the skin.

In the provided exercise, heat loss by convection and radiation is mentioned as a combined rate. This simplification is common in problems involving human heat transfer, as distinguishing between the two can be complex and depends on numerous environmental variables. However, both mechanisms are essential for understanding and managing heat loss in different climates, during exercise, or in various medical applications.

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Most popular questions from this chapter

Hot water \(\left(c_{p}=4.179 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) flows through a 200-m-long PVC \((k=0.092 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) pipe whose inner diameter is \(2 \mathrm{~cm}\) and outer diameter is \(2.5 \mathrm{~cm}\) at a rate of \(1 \mathrm{~kg} / \mathrm{s}\), entering at \(40^{\circ} \mathrm{C}\). If the entire interior surface of this pipe is maintained at \(35^{\circ} \mathrm{C}\) and the entire exterior surface at \(20^{\circ} \mathrm{C}\), the outlet temperature of water is (a) \(39^{\circ} \mathrm{C}\) (b) \(38^{\circ} \mathrm{C}\) (c) \(37^{\circ} \mathrm{C}\) (d) \(36^{\circ} \mathrm{C}\) (e) \(35^{\circ} \mathrm{C}\)

Hot air is to be cooled as it is forced to flow through the tubes exposed to atmospheric air. Fins are to be added in order to enhance heat transfer. Would you recommend attaching the fins inside or outside the tubes? Why? When would you recommend attaching fins both inside and outside the tubes?

The boiling temperature of nitrogen at atmospheric pressure at sea level ( 1 atm pressure) is \(-196^{\circ} \mathrm{C}\). Therefore, nitrogen is commonly used in low-temperature scientific studies since the temperature of liquid nitrogen in a tank open to the atmosphere will remain constant at \(-196^{\circ} \mathrm{C}\) until it is depleted. Any heat transfer to the tank will result in the evaporation of some liquid nitrogen, which has a heat of vaporization of \(198 \mathrm{~kJ} / \mathrm{kg}\) and a density of \(810 \mathrm{~kg} / \mathrm{m}^{3}\) at 1 atm. Consider a 3-m-diameter spherical tank that is initially filled with liquid nitrogen at 1 atm and \(-196^{\circ} \mathrm{C}\). The tank is exposed to ambient air at \(15^{\circ} \mathrm{C}\), with a combined convection and radiation heat transfer coefficient of \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The temperature of the thin-shelled spherical tank is observed to be almost the same as the temperature of the nitrogen inside. Determine the rate of evaporation of the liquid nitrogen in the tank as a result of the heat transfer from the ambient air if the tank is \((a)\) not insulated, \((b)\) insulated with 5 -cm-thick fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and (c) insulated with 2 -cm-thick superinsulation which has an effective thermal conductivity of \(0.00005 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

A \(2.5 \mathrm{~m}\)-high, 4-m-wide, and 20 -cm-thick wall of a house has a thermal resistance of \(0.0125^{\circ} \mathrm{C} / \mathrm{W}\). The thermal conductivity of the wall is (a) \(0.72 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (b) \(1.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (c) \(1.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (d) \(16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (e) \(32 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)

Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially that of the surrounding air, i.e. \(20^{\circ} \mathrm{C}\). Its width is \(5.0 \mathrm{~cm}\); thickness is \(1.0 \mathrm{~mm}\); thermal conductivity is \(200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\); and base temperature is \(40^{\circ} \mathrm{C}\). The heat transfer coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Estimate the fin temperature at a distance of \(5.0 \mathrm{~cm}\) from the base and the rate of heat loss from the entire fin.

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