Physics Including Human Applications 238 Chapter 11 Thermal Transport GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions Define the following terms, and use them in an operational definition: conduction radiation convection Energy Transfer Problems Solve numerical problems that involve a transfer of energy, temperature gradients, and conduction, convection, or radiation. Living System Thermal Properties Explain basic thermal effects in living systems. PREREQUISITES Before beginning this chapter, you should have achieved the goals of Chapter 9, Transport Phenomena, and Chapter 10, Temperature and Heat.
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Chapter 11 Thermal Transport GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions Define the following terms, and use them in an operational definition: conduction radiation convection Energy Transfer Problems Solve numerical problems that involve a transfer of energy, temperature gradients, and conduction, convection, or radiation. Living System Thermal Properties Explain basic thermal effects in living systems. PREREQUISITES Before beginning this chapter, you should have achieved the goals of Chapter 9, Transport Phenomena, and Chapter 10, Temperature and Heat.
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Chapter 11 Thermal Transport
11.1 Introduction In a time of energy scarcity, all of us become more concerned about the ways we can conserve energy. Various community programs have been started to teach us how to use energy more efficiently. In particular, agencies of government have encouraged us to improve the thermal transport properties of our homes. What can we do to reduce the loss of heat from our homes in the winter? Or to prevent the overheating of our homes in the summer? In this chapter we will examine the processes of thermal transport. 11.2 Conduction We have already discussed the flow of heat through solid objects in Chapter 9. You may recall that we found that the rate at which heat travels along the handle of a silver spoon is related to the temperature gradient (see Section 9.2). The process by which heat energy travels from one part of a solid object to another is called conduction. Conduction of heat energy takes place in an object only when different parts of the object have different temperatures. The direction of heat flow is always from the point of higher temperature to that of lower temperature. Conduction involves a transfer of energy within a substance without the material itself being in motion. Conduction is the primary process by which heat energy travels through solids. The thermal conductivity κ of a material is defined as the flow of heat energy per unit time per unit area per unit temperature gradient, in equation form, κ = - H/A(ΔT/Δx) (11.1) where H is the number of joules of energy transferred in one second, A is the area in square meters, and ΔT/Δx is the temperature gradient in degrees Celsius per meter. The negative sign is introduced because H is positive and in the opposite sense to the temperature gradient ΔT/Δx. The expression for the flow of heat energy in watts or J/sec is given by H = -κA ((ΔT/Δx) (11.2) For most materials the thermal conductivity is a function of temperature, but the variation with temperature is small for solids and for our purposes may be neglected.
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EXAMPLES
1. Compare the heat energy loss by conduction from a brick house of 15 cm wall thickness and from a wooden house of 5-cm wall thickness. Assume the temperature difference between the inside and the outside is 20øC, the house is 6 m x 8 m x 4 m tall, and the heat energy lost through the floor is zero. Brick HB = -κA (ΔT/Δx) = -(0.63)watts/m-øC[2(6.0)(4.0) + 2 (8.0)(4.0) + (6.0)(8.0)]m2 x 20øC/0.15 m = 1.3 x 104 watts Wood Hw = (0.15) watts/møC (160 m2) (20øC/0.05 m) = 9.6 x 103 watts
2. During the energy crisis of 1973 many homeowners were encouraged to put storm windows on their homes. Does that really make any difference? Compare the heat loss due to conduction for a single window pane and for a storm-window pane with a dead air space between two glass panes, where the dead air space is twice as thick as a pane of glass (see Figure 11.1). Let us use the following symbols: T1 is the inside temperature and T2 is the outside temperature, soT1 > T2, κ1 is thermal conductivity of glass, κ2 is thermal conductivity of dead air, x1 is glass thickness, x2 is air thickness (= 2x1), and A is the area of window. At equilibrium the heat flow through all interfaces must be equal, the heat flow through the inside pane of glass = heat flow through the dead air space = heat flow through the outer pane of glass -κ1A ((T1 - T')/x1) = -κ2A ((T' - T")/x2) = -κ1A ((T" - T2)/x1) where T' is the temperature at the outside of the inner pane of glass andT" is the inside of the outer pane of glass. Eliminating T' and T", we get the following equation for heat transfer through the thermal pane, H = [-A (T1 - T2)]/(x1/κ1 + x2/κ + x1/κ1) (11.3) This result can be generalized for N layers, H = -A (T1 -T2)] / [Σi=1
N (xi/κi) (11.4) where κi is the thermal conductivity of the ith layer of thickness xi, and (T1 - T2) is the temperature difference across the N layers.
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For the storm window we have: Hsw = [-A (T1 -T2)] / [2x1/κ1 + x2/κ2] = [-A (T1 -T2)] /[2x1/0.84 + 2x1/0.024] For a single glass pane we have: Hg = [-A (T1 -T2)] / [x1/κ1] = [-A (T1 -T2)] / [x1/0.84] The ratio of storm window heat loss to glass pane heat loss is given by (Hsw/Hg) = (x1/.84) / (2x1/.84 + 2x1/ .024) = 1.2/(2.4 + 83) = 1/71 In this case, the storm window reduces the loss of thermal energy by a factor of 71! 3. A skier has a body surface area of 2.0 m2 and wears clothing 1.0 cm thick. The thermal conductivity of the dry clothing is 4.0 x 10-2 watts/m-øK. Assume that the skier's skin temperature is 33øC and the outer clothing surface temperature is 0.0øC. We wish to find the heat loss through the dry clothing and compare it with that through wet clothing (which has a thermal conductivity equal to that of water = 59 x 10-2 watts/m-øK. Hd = κdA(Ts -To)/x, = -4 x 10-2 x (33 - 0) x 2/10-2 watts = -260 watts (heat loss through dry clothing) Hw = - κwA(Ts -To)/x = -59 x 10-2 x (33 - 0) x 2/10-2 watts = -3900 watts (heat loss through wet clothing) This example illustrates why one feels cooler in wet clothing.
11.3 Convection You have observed a pan of water heating on a stove, and you have probably noticed that the water seemed to be in motion. Also you have observed the trail of the smoke from a lighted cigarette. These are examples of the transfer of heat energy by convection, a process in which the heated, or higher energy, matter moves. The transfer of heat by convective circulation in a liquid or gas is generally brought about by a difference in density. Liquids and gases, when heated, expand and become less dense. The less dense material is continually displaced by the fluid of greater density. This motion is known as convection currents, and heat energy is transferred from one location to another by the motion of the matter with higher internal, or heat, energy. Convection is an important factor in the design of insulation systems. In order to minimize convection losses, it is necessary to break up open spaces so that the circulation of matter is impeded. For this reason the insulating material used in the walls of a refrigerator or in walls of a house is porous. Insulating materials are not only poor conductors but they contain many small air spaces. No effective long-range convection currents can be set up in the insulating materials. A detailed mathematical theory of heat convection is quite complicated, but we will discuss the loss of heat energy from the human body by convection in an introductory quantitative way.
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Your body has a relatively stable interior temperature, Tb. If you are located in air of temperature Ta which is different from your body temperature there will be a flow of heat by conduction through your skin Ts to the air (Figure 11.2). Using our previously developed equations, we can write statements for the heat current through the skin and through the air as follows:
H = (-κs/Δx) (Ts - Tb) A where κs is the coefficient of thermal conductivity of the skin with thickness Δx. H = (-κa/Δx) (Ta - Ts) A where κa is the coefficient of thermal conductivity for still air with thickness Δx. We can eliminate Ts from these equations as we did in the example of the storm window, and we obtain H = [-κsκa/(κs+κa)] • [(Tb - Ta)/Δx]A (11.5) Notice that the heat energy flow across the interface is proportional to the total temperature drop from the interior body temperature to the temperature of the air, Tb - Ta. The coefficient of thermal conductivity of still air is small; so the flow of heat energy through perfectly still air is rather low. One of the functions of warm clothes is to provide traps for air to prevent the movement of air near the body and hence to provide good insulation for the body. However, if air is forcibly circulated, or if there is a considerable amount of natural air movement, the rate of energy flow increases because of convection (Figure 11.3). !
Right at the skin-air interface, we assume that the flow of heat energy is proportional to the total temperature difference between your skin and the air, Ts - Ta. We can write an expression for the heat energy flow by force convection (no stagnant air layer at the surface) as H = -κ' (Ts -Ta) (11.6) where κ' is the convection coefficient of the object, which depends upon the nature and area of the interface, the speed of the air movement, the density and viscosity of air, and so on. The value of κ' is usually determined experimentally for a given system. Under normal indoor conditions, approximately 30 percent of the total heat energy lost from the human body is through convection. As the speed of the air past the body surface increases, as for a bicycle rider, for example, the fraction of heat lost by convection increases.
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EXAMPLE The convection coefficient of a clothed human body at rest is about 2 watts/m-øC. When the air velocity is 5 m/sec with respect to the body, as for a bicycle rider, the convection coefficient increases to 28 watts/m-øC. Compare the rates of heat loss in these two cases assuming the same body and environment temperature.
An empirical rule, known as Newton's law of cooling, was developed to predict the rate at which hot objects are cooled by convection. This rule has the same form as Equation 11.6 and states in mathematical form that the rate of cooling is proportional to the difference between the temperature of the object and the temperature of its surroundings. It can be shown, using the methods of calculus (see Section 11.7), that the temperature of the cooling object can be expressed as an exponential function of the time of cooling, T = To +ΔTexp(-κ't/C) (11.7) where κ' is the convection coefficient, C is the heat capacity of the object, ΔT is the initial temperature difference between the environment and the object, T0 is the temperature of the environment, and T is the temperature of the object at a time t. EXAMPLE A boiling hot cup of coffee (heat capacity 72 J/øC) in a room of 20øC cools down to a
temperature of 60øC in one minute. What is the convection coefficient of this system? How long will it take for the coffee to become lukewarm (48øC)? To find the convection coefficient, use Equation 11.7 where all the quantities are known except κ', T = To +ΔTexp(-κ't/C) 60 = 20 + (100 - 20) exp(-κ'(60 sec)/72 J/øC) 40 = 80 exp( 0.833κ'sec-øC/J) writing 0.50 as exp( ln0.50), we obtain, 0.50 = exp( -0.69) = exp(0.83κ') so 0.83 κ' = 0.69 J/sec-øC κ' = 0.83 J/sec-øC
Notice that the initial temperature difference of 80ø has dropped to a temperature difference of 40ø in one minute. It is the property of exponential functions to have constant half-times; that is, the time required for the value of the exponential function to change by a factor of two is constant. Hence, the coffee will have a temperature of 40øC (a temperature difference of 20ø) after 2 minutes, 30øC after 3 minutes; 25øC after 4 minutes; 22.5ø after 5 minutes, and so on.
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11.4 Radiation You know that heat energy is transferred from the sun to the earth. This method of transfer is called radiation. We say that the sun is radiating energy as electromagnetic waves. This type of radiation does not require a material medium for its propagation. Since it is an example of interaction-at-a-distance, we propose that the energy is transferred to the earth from the sun by means of a field (see Chapter 1). Any object at a temperature above absolute zero emits energy from its surface. This energy is in the form of electromagnetic waves and is called radiant energy. The waves travel with the velocity of light. They may be absorbed by the medium in which they travel or by the object which they strike. When this radiant energy is absorbed it becomes heat energy. An object that absorbs this radiant energy, converts the radiant energy into internal, or heat, energy. Experiments have shown that good absorbers are also good radiators. So an ideal radiator is an object that absorbs all of the radiation that falls upon it. It is called anideal black body. No perfect black body is known, but we have some good approximations such as objects whose surfaces are coated with lamp black. Objects with bright shiny surfaces are poor radiators.
The law that expresses the total energy radiated by a black body was originally stated by Stefan on the basis of experimental measurements from a black-body cavity. Later Boltzmann derived the equation from thermodynamic theory. The rate P at which energy is radiated by a black body is given by P = σAT4 (11.8) where P is expressed in watts, A is the area of the radiator in square centimeters, and T is the Kelvin temperature, and σ has a value of 5.70 x 10-12 watt/cm2-øK4. This is known as the Stefan-Boltzmann law. Anything other than a black body will radiate at a lower rate. The distribution of the energy among the different wavelengths of radiation is characteristic of the temperature of the emitting surface (Figure 11.4). From this distribution you can explain the saying, "White hot is hotter than red hot."
Let us consider a greenhouse. Window glass is transparent for visible radiation but absorbs heat radiation. The glass roof transmits the visible and near infrared radiations from the sun (T ~ 6000øK). These rays are converted to heat when they are absorbed by objects in the greenhouse. The absorbing objects are heated and become radiators. Their radiation is characteristic of their temperature (about 360øK). Because the glass is opaque to this radiation, this energy does not get out but is reflected back into the greenhouse. So a greenhouse serves as an energy trap.
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11.5 Thermal Effects in Living Systems A living system is constantly interacting thermally with its environment. Environmental equilibrium temperatures on earth are maintained by radiation from the sun. (The solar power reaching the earth is approximately 1.4 x 103 watts/m2.) Living systems maintain a steady-state equilibrium with their environment by adjusting their metabolic energy production to balance their heat transfer to the environment. The basal metabolic rate is the heat production under normal resting conditions. Sleeping organisms operate at levels below basal rates, and active systems operate at higher metabolic rates. Only a limited range of temperatures will support living systems as we currently know them. Temperature changes can alter the form of proteins in living systems. Such alterations may result in drastic changes in large protein chains. When this occurs the proteins are no longer capable of performing their biological function (such proteins are said to be thermally denatured). For many proteins the range of temperatures from 10ø to 20øC is the range for maximum stability. Much lower or higher temperatures produce denatured proteins. In all animals the metabolic rate increases as the body temperature increases. In humans a 1øC temperature rise produces about a seven percent increase in the basal metabolic rate. Different systems have maximum efficiency for metabolic processes at different temperatures. Those animals that show a constant body temperature are calledhomeotherms(warm-blooded animals). Animals not possessing the necessary thermoregulation system for a constant body temperature are called poikilotherms (cold-blooded animals). The temperature of poikilotherms varies with their environmental temperature. Reptiles are examples of poikilotherms. It seems clear that homeotherms have an evolutionary advantage as they can function independent of significant environmental temperature changes. It has also been suggested that the development and functioning of a complex brain depends on the maintenance of a constant brain temperature. In humans the core temperature for internal organs is 37øC while the skin temperature is about 33øC. For a human with a surface area of 2 m2 the basal metabolic rate (BMR) is about 90 watts. Exercise can increase the metabolic rate by a factor of 10. The conditions affecting the heat transfer to the environment are important factors in determining the metabolic rate. Heat is transferred to the skin from the core of the body by conduction and by forced convection of the blood flowing from the core to the peripheral blood vessels. At high external temperatures or high metabolic rates the human body compensates by increasing its rate of energy loss. This is done by increasing the peripheral blood circulation, dilating the blood vessels (vasodilation), and by increasing evaporation cooling. Vasodilation can increase the effective heat transfer by a factor of eight and under optimum conditions sweating can produce a cooling rate of approximately 700 watts. At low temperatures the human system acts to reduce heat transfer. Vasoconstriction reduces blood flow to the surface and thereby reduces heat flow. Energy production is increased for short times by shivering. For longer periods of time, enzyme activity increases metabolic activity in order to maintain equilibrium. The control element for the human body thermoregulator is in the hypothalamus, located at the base of the brain. The hypothalamus is connected by the nervous system to the thermal detectors on the body surface. The detective work involved in solving the mystery of the human temperature regulation is an example of success in biophysical research.
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Svante A. Arrhenius made one of the most successful theoretical formulations for chemical reaction rates. He suggested that if chemical reactions take place because of collision interactions among molecules, then the rates should be governed by the Maxwell-Boltzmann distribution of energy for molecules in thermal equilibrium at absolute temperature T, a law of classical physics. This distribution can be expressed in equation form as follows: P ∝ exp(- E/kT) where P is the probability that a system has energy E when it is in thermal equilibrium at an ABSOLUTE TEMPERATURE T and where k is Boltzmann's constant, which has a value of 1.38 x 10-23 J/øK. The reaction rate equation derived by Arrhenius is: λ = Aoexp(- Ea/kT) (11.9) where Ea is the activation energy for the reaction, λ is the rate of the reaction, and Ao is a constant. It has been shown that this same rate equation applies to the activity of biological systems such as fruit flies, guppies, and brine shrimp. The experimental set up consists of a test chamber that is divided into two equal parts separated by a transparent dividing partition in which there is a hole that can be opened. All samples are initially placed in one side of the test chamber, and the hole is opened. The number of samples reaching the second chamber is recorded as a function of time. If the samples are treated as molecules in a kinetic theory model, it can be shown that the following equation results for the number of samples in the second chamber as a function of time: N2 = (N0/2) (1 –exp(-λt)) Or ln [N0 / (N0 - 2N2)] = λt (11.10) where N0 is the original number of samples placed in side 1, N2 is the number of samples in side 2 at time t, and λ is the reaction rate (number/sec passing through the hole). Using this equation and solving for λ, we can then use the Arrhenius equation to determine an activation energy for the activity involved. In a graph of the natural logarithm of λ versus 1/kT, the slope is equal to the activation energy. (See footnote 1.) This approach suggests an interesting way to study the effects of such variables as chemicals, light, and magnetic fields on the activity of biological systems. For example consider the question, "Do mosquitoes find their targets through chemical sensing or thermal sensing?" Design an experimental system, and start counting mosquitoes. EXAMPLE Natural convection from the human body is found to obey the following heat energy flow equation: H = -κ(Ts - Ta)1.25 watts/m2. We want to compare the natural convection of heat energy loss of a nude person with skin temperature of 33øC in a room at 18øC to the heat loss in the same room maintained at 23øC. H18 = -κ (33 - 18)1.25 H23 = -κ (33 - 23)1.25 Thus we have H18/H23 = (15/10)1.25 = 1.66 Thus we have H18/H23 = (15/10)1.25 = 1.66 The person loses heat 1.66 times faster in a room at 18øC than in the same room at 23øC.
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11.6 Microwave Cooking A recent development in the use of electromagnetic radiation as a source for heat energy is microwave cooking. We offer below two different explanations for how microwaves cook. The first explanation is from a microwave oven manufacturer's cookbook. The second is from an article that evaluated the performance of various microwave ovens.
HOW MICROWAVES COOK Traditionally, food heats and cooks because of molecular activity caused by a gas flame, burning wood and charcoal, or electricity converted to heat energy. This intense heat must be applied to the bottom of a pan of food or used to surround food in an oven with hot air. If food comes in direct contact with these traditional heat sources, it burns before it cooks through. Now electrical energy can be converted to microwave energy by means of an electron tube called a magnetron. This tube is inside the microwave oven and sends microwaves directly into food. Microwaves are classified as electromagnetic waves of a non-ionizing frequency. Microwaves travel directly to food without heating the air or the recommended cooking dishes. The cooking process speeds up because it starts as soon as the oven is turned on. Microwaves move directly to food because they are attracted to the fat, sugar and liquid or moisture molecules, causing them to vibrate at a fantastically fast rate. This vibration is heat energy. The vibrating molecules bump and rub others, start these molecules vibrating and set up a chain reaction that moves from the outside edges, where microwaves first come in contact with food, toward the center--cooking as it goes. This chain reaction is called conduction. The molecule vibration, or cooking, continues for several minutes after food comes from a microwave oven and is taken into consideration in microwave recipes. Microwaves' specific attraction for moisture, fat and sugar in molecules, plus the fast molecule vibration rate this causes, results in the amazing speed of microwave cooking. (See footnote 2.) MICROWAVE COOKING "Microwaves reflect from metal objects and pass right through glass, paper and some plastic objects... but are absorbed by foods and
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converted to heat. The waves are not hot; they create heat by agitating molecules in materials that are capable of absorbing them. In addition to food, that includes human flesh and bone. The drawing (Figure 11.5) shows how the microwaves issue from an electronic vacuum tube, called a magnetron, flow through a metal tube, or wave guide, and are dispersed around the oven cavity by a rotating metal paddle, or stirrer. As the microwaves penetrate the food, more energy reaches the interior portion immediately in a microwave oven than in a conventional oven, so that cooking, overall, can be much faster. But the actual time required will depend on the type and quantity of food." (See footnote 3). Using the concepts you have studied, evaluate these two explanations of cooking in microwave ovens. Which explanation seems more vivid? ENRICHMENT 11.7 Newton's Law of Cooling Let us begin with the equation for the flow of heat energy by forced convection, Equation 11.6 H = -κ' (T - To) (11.6) where H is the flow of heat in watts, T is the temperature of the object and T0 is the temperature of the surroundings. In the notation of calculus the flow of heat is the time derivative of the heat energy of the system, H = dQ/dt where Q is the internal energy of the object and Q = mcT where m is the mass of the object, c is the specific heat of the object and T is the temperature of the object (from Chapter 10). So we can rewrite Equation 11.6 in the calculus notation to obtain the standard form of Newton's law of cooling, mc (dT/dt) = -κ' (T - To) (11.11) Let us use the technique of separation of variables and then integration to find an expression for the temperature of the object as a function of time. dT/( T - To) = - (κ' dT)/mc ln (T - To) = (-κ't)/(mc) + constant (T - To) = (constant) exp(- κ't/mc) (11.12) Since at t = 0, the object was ΔT hotter than the environment, we obtain Equation 11.7 T = To + ΔTexp( -κ't/mc) Remember mc = C (the heat capacity of body).
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SUMMARY Use these questions to evaluate how well you have achieved the goals of this chapter. The answers to these questions are given at the end of this summary with the section number where you can find the related content material. Definitions Write the correct word or phrase in each blank.
1. When heat energy is transferred through a solid object it occurs by a process called ______.
2. Metals have a larger ______ than gases and transfer heat energy by ______. 3. The movement of higher temperature, lower density portions of fluids gives rise
to a process of heat energy transfer called ______. 4. In the "zero gravity" of an orbiting space craft it has been found that ______ is no
longer an important process of heat transfer because ______. 5. In contrast to the other energy transfer processes ______ can occur in the absence
of a material medium. Energy Transfer Problems
6. How many watts would be conducted through a copper sheet 25 x 25 cm square and 1 mm thick with a temperature difference of 200øC?
Living System Thermal Properties Write the correct word or phrase in each blank.
7. Living systems that are able to maintain a constant temperature are called ______. 8. Such systems are able to control the rate of their internal ______ by their precise
control of ________. 9. The _______ is the control element for the thermoregulation of the human body.
Answers
1. conduction (Section 11.2) 2. coefficient of thermal conductivity, conduction (Section 11.2) 3. convection (Section 11.3) 4. convection, differences in density no longer result in fluid motion (Section 11.3) 5. radiation (Section 11.4) 6. 4.75 x 106 watts (Section 11.2) 7. homeotherms (Section 11.5) 8. chemical reactions, temperature (Section 11.5) 9. hypothalamus (Section 11.5)
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ALGORITHMIC PROBLEMS Listed below are the important equations from this chapter. The problems following the equations will help you learn to translate words into equations and to solve single-concept problems. Equations H = -κA ((ΔT/Δx) (11.2) H = -A (T1 -T2)] / [Σi=1
N (xi/κi) (11.4) H = -κ' (Ts -Ta) (11.6) T = To +ΔTexp(-κ't/C) (11.7) P = σAT4 (11.8) λ = Aoexp(- Ea/kT) (11.9) Problems
1. A temperature gradient of 40øC/cm is maintained across an aluminum rod 4 cm2 in area. What is the rate of energy transfer by conduction?
2. If the absolute temperature of an object is doubled, by what factor does the rate of energy radiated from the object change?
3. Compare the rate at which energy is radiated by an object at room temperature (27øC) to the rate at which energy is radiated by the object at the temperature of liquid helium (4øK).
4. A hot object of 80øC is placed in a 0øC environment where its energy loss occurs primarily by convection. After one minute the temperature of the object is 60øC; what will the temperature of the object be in 5 minutes? κ'/C = 0.288/min for this object
Answers 1. 320 watts 2. 16 3. 3.16 x 107 4. 19øC
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EXERCISES These exercises are designed to help you apply the ideas of a section to physical situations. When appropriate the numerical answer is given in brackets at the end of each exercise. Section 11.2
1. Compare the heat loss by conduction for 5.0 mm of dead air plus 5.0 mm of clothing to that for 1.0 cm of clothing. Assume the temperature difference between the skin and environment to be 40øC. [1.3 times as much heat lost for clothing as for combination]
Section 11.3
2. Under normal forced air convection conditions, the convection coefficient for the human body in air is 13.95 watts/m-øC. If the convective heat current density loss from a body is 60.0 watts/m2 for resting, find the temperature difference between the skin and room air. [4.30øC]
Section 11.4
3. If the absolute temperature of a radiating body is increased by 50 percent, how does the rate of radiation increase? [Pnew/Pold = 5.1]
PROBLEMS Each of the following problems may involve more than one physical concept. The numerical answer to the problem is given in brackets at the end of the problem.
4. Compare wood and glass as insulating materials. 5. Three metal rods - copper, aluminum, and brass - each 6 cm long and 2 cm in
diameter are placed end to end (Figure 11.6). The free ends of copper and brass are held at temperatures of 100øC and 0øC respectively. Assume that the thermal conductivity of copper is two times that of aluminum, and that of aluminum is two times that of brass. What is the temperature at the copper-aluminum junction and at the aluminum-brass junction when the system reaches equilibrium condition? [T Cu-Al = 600/7 ≈ 86øC,TAl-Br = 400/7 ≈ 57øC]
6. Analyze the heating system of your home in terms of conduction, convection, and
radiation.
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7. A house has insulating brick walls of 26 cm thickness and a total area of 240 m2. If the difference in temperature between the inside and outside is 20øC, how much natural gas, which has a heat of combustion of 10,000 cal/g, will it take per day to maintain a constant temperature? [5600 g]
8. Given the problem of designing a cold storage container for dry ice using styrofoam of constant thickness, compare the rate of heat flow into spherical, cylindrical (d = h), and cubical containers of equal volume when containers are used in identical conditions. [For identical conditions the heat flow will be proportional to the surface area of the container. For a volume of (4π/3) m3, the sphere will have a radius of 1 m and a surface area of 4π m2. The cylinder will have a radius of 0.88 m, a height of 1.76 m, and an area of 4.65π m2. The cube will have a side length of 1.61 m, and an area of 15.54 m2. Hence the heat flow will be the smallest into the spherical container, next smaller into the cylinder, and greatest into the cube in the ratio of 1:1.16:1.24.]
9. A paramedic team rushes to the scene of an accident and finds a corpse of temperature 26.2øC, five minutes later the corpse temperature is 23.8øC. If the environment temperature is 22.0øC, the normal body temperature is assumed to be 37.0øC, and the rate of heat energy loss assumed to be forced convection cooling, what time elapsed between the death of the person and the arrival of the paramedic team? [7.46 min]
10. Four identical glass, double-walled containers are treated in different ways by evacuating the air from between the walls or not, and by silvering the glass walls or not. Then each container is nearly filled with hot water and a cork stopper is placed in each container. Discuss the rate of heat energy loss from each container (Figure 11.7).
11. Assume a space traveler finds three other planets about the same size as the Earth but with helium, neon, and oxygen atmospheres. Compare the rates of heat conduction from these planets to that from the Earth. What additional assumptions must you make?
Footnotes 1) O. A. Runquist, C. J. Creswell, J. T. Head-Burgess,Chemical Principles: A Programmed Text, 2nd ed. (Minneapolis, Minn.: Burgess, 1974), p. 383. 2) Variable Power Microwave Cooking, Litton Systems, Inc.(New York: Van Nostrand Reinhold, 1975), p. 6. 3) Consumer Reports, Consumers Union of United States, Inc., 38, Apr. 1973, p.221.
