/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 8\. The thermal conductivities o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

8\. The thermal conductivities of human tissues vary greatly. Fat and skin have conductivities of about \(0.20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(0.020 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively, while other tissues inside the body have conductivities of about \(0.50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assume that between the core region of the body and the skin surface lies a skin layer of \(1.0 \mathrm{~mm}\), fat layer of \(0.50 \mathrm{~cm}\), and \(3.2 \mathrm{~cm}\) of other tissues. (a) Find the \(R\) -factor for each of these layers, and the equivalent \(R\) -factor for all layers taken together, retaining two digits. (b) Find the rate of energy loss when the core temperature is \(37^{\circ} \mathrm{C}\) and the exterior temperature is \(0^{\circ} \mathrm{C}\). Assume that both a protective layer of clothing and an insulating layer of unmoving air are absent, and a body area of \(2.0 \mathrm{~m}^{2}\)

Short Answer

Expert verified
The \(R\)-factors for the skin, fat, and other tissues and the equivalent \(R\)-factor for all layers together needs to be computed first. Once the total resistance is found, it allows to calculate the rate of energy loss by subtracting the exterior temperature from the core body temperature and dividing by the total resistance.

Step by step solution

01

Find the Resistance factor for each layer

The resistance \(R\) for heat flow through a layer of material can be calculated using the formula \(R = L / (k \cdot A)\), where \(L\) is the thickness of the layer, \(k\) is the thermal conductivity of the material, and \(A\) is the area through which heat is flowing. For the skin layer: \(R_{skin} = (0.1 \, \mathrm{cm}) / (0.020 \, \mathrm{W/m\cdot K} \cdot 2.0 \, \mathrm{m}^{2})\). For the fat layer: \(R_{fat} = (0.50 \, \mathrm{cm}) / (0.20 \, \mathrm{W/m\cdot K} \cdot 2.0 \, \mathrm{m}^{2})\). For the other tissues layer: \(R_{tissues} = (3.2 \, \mathrm{cm}) / (0.50 \, \mathrm{W/m\cdot K} \cdot 2.0 \, \mathrm{m}^{2})\).
02

Calculate the total Resistance factor

The total resistance \(R_{\mathrm{total}}\) for the heat flow across multiple layers is simply the sum of the resistances of the individual layers. Hence, \(R_{\mathrm{total}} = R_{skin} + R_{fat} + R_{tissues}\).
03

Compute the rate of energy loss

The rate of energy loss \(Q\) due to heat conduction can be determined using the formula \(Q = (T_{\mathrm{hot}} - T_{\mathrm{cold}}) / R_{\mathrm{total}}\), where \(T_{\mathrm{hot}}\) is the core temperature, \(T_{\mathrm{cold}}\) is the exterior temperature and \(R_{\mathrm{total}}\) is the total resistance computed in the previous step. Hence, \(Q = (37^{\circ} \mathrm{C} - 0^{\circ} \mathrm{C}) / R_{\mathrm{total}}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Heat Transfer
Understanding the way heat moves is crucial to grasp a range of scientific concepts, from baking a cake to keeping our bodies warm. Heat transfer is the process by which thermal energy moves from one place to another. It occurs via three primary mechanisms: conduction, convection, and radiation.

In our exercise, conduction is the primary focus, which is the transfer of heat through materials. Imagine holding a hot cup of coffee. The heat travels from the hot lower part of the cup through the material to your hand. That's conduction at work. It happens due to the movement and collision of particles within an object; the faster-moving (hotter) particles collide with the slower-moving (colder) ones, passing on some of their energy.

The rate at which heat is conducted depends on the thermal conductivity of the material (\( k \)), which is measured in watts per meter-kelvin (\( W/m \times K \)). Materials with higher thermal conductivity transfer heat more quickly. In our body, different tissues conduct heat at different rates, hence, they have varying thermal conductivities.
Thermal Resistance
Just as electrical resistance hinders the flow of electricity, thermal resistance impedes the flow of heat. This property is a measure of how difficult it is for heat to pass through a material. It's directly tied to the concept's 'R-value' or 'R-factor', which comes into play in the textbook exercise.

Calculate the thermal resistance (\( R \) for heat flow through a layer with the formula \( R = L / (k \times A) \), where \( L \) is the thickness, \( k \) is the thermal conductivity, and \( A \) is the area of the material. If you pile on several layers, like clothing or insulation in a wall, you just add up their R-values to get the total thermal resistance (\( R_{total} \)).

In our problem, we investigate different layers of human tissue, each with its unique R-factor. The total R-factor is the sum of these, illustrating how the body's natural layers contribute to retaining heat. One crucial aspect to remember is that the R-factor is typically used to compare insulative properties—the higher the R-factor, the better the insulation.
Energy Loss in Humans
Energy loss in humans is essentially heat escaping from the body into the environment, a crucial aspect of thermodynamics relevant to our survival. Our core body temperature is tightly regulated, but when the external temperature is lower, heat naturally flows out of our body to equalize with the cooler environment. This flow can be fast or slow, depending on the resistance to heat transfer, which is where our exercise ties in.

Factors influencing energy loss include fat and skin layers, which provide insulation, and the surrounding temperature. Fat, with a lower thermal conductivity, acts as a good insulator. The more fat, the slower the heat loss. But if we're in a cold environment with minimal clothing or shelter—as the exercise assumes—the rate of energy loss ramps up.

Using the calculated R-factor, we can find the rate of energy loss by considering the temperature gradient and the total thermal resistance. This helps us understand how quickly we'd lose heat in a cold environment, a measurement that's crucial for designing effective clothing, heating systems, and even medical interventions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(200-\mathrm{g}\) aluminum cup contains \(800 \mathrm{~g}\) of water in thermal equilibrium with the cup at \(80^{\circ} \mathrm{C}\). The combination of cup and water is cooled uniformly so that the temperature decreases by \(1.5^{\circ} \mathrm{C}\) per minute. At what rate is energy being removed? Express your answer in watts.

An unknown substance has a mass of \(0.125 \mathrm{~kg}\) and an initial temperature of \(95.0^{-} \mathrm{C}\). The substance is then dropped into a calorimeter made of aluminum contain. ing \(0.285 \mathrm{~kg}\) of water initially at \(25.0^{\circ} \mathrm{C}\). The mass of the aluminum container is \(0.150 \mathrm{~kg}\), and the temperature of the calorimeter increases to a final equilibrium temperature of \(32.0^{\circ} \mathrm{C}\). Assuming no thermal energy is transferred to the environment, calculate the specific heat of the unknown substance.

A Styrofoam box has a surface area of \(0.80 \mathrm{~m}^{2}\) and a wall thickness of \(2.0 \mathrm{~cm}\). The temperature of the inner surface is \(5.0^{\circ} \mathrm{C}\), and the outside temperature is \(25^{\circ} \mathrm{C}\). If it takes \(8.0 \mathrm{~h}\) for \(5.0 \mathrm{~kg}\) of ice to melt in the container, determine the thermal conductivity of the Styrofoam.

A "solar cooker" consists of a curved reflecting mirror that focuses sumlight onto the object to be heated (Fig. \(P 11.67\) ). The solar power per unit area reaching the Farth at the location of a \(0.50-\mathrm{m}-\) diameter solar cooker is \(600 \mathrm{~W} / \mathrm{m}^{2}\). Assuming \(50 \%\) of the incident energy is converted to thermal energy, how long would it take to boil away \(1.0 \mathrm{~L}\) of water initially at \(20^{\circ} \mathrm{C} ?\) (Neglect the specific heat of the container.)

ecp Liquid nitrogen has a boiling point of \(77 \mathrm{~K}\) and a latent heat of vaporization of \(2.01 \times 10^{\circ} \mathrm{J} / \mathrm{kg}\). A \(25-\mathrm{W}\) electric heating element is immersed in an insulated vessel containing \(25 \mathrm{~L}\) of liquid nitrogen at its boiling point. (a) Describe the energy transformations that occur as power is supplied to the heating element. (b) How many kilograms of nituogen are boiled away in a period of \(4.0\) hours?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.