Ch13. The Transfer of Heat

Convection Convection is the process in which heat is carried from place to place by the bulk movement of a fluid. Convection currents are set up when a pan of water is heated.

Conceptual Example 1. Hot Water Baseboard Heating and Refrigerators

Hot water baseboard heating units are frequently used in homes, where they are mounted on the wall next to the floor. In contrast, the cooling coil in a refrigerator is mounted near the top of the refrigerator. The locations for these heating and cooling devices are different, yet each location is designed to maximize the production of convection currents. Explain how. (a) The air above the baseboard unit is heated, like the air above a fire. Buoyant forces from the surrounding cooler air push the warm air upward. Cooler air near the ceiling is displaced downward and then warmed by the baseboard heating unit, leading to the convection current. Had the heating unit been located near the ceiling, the warm air would have remained there, with very little convection to distribute the heat.

(b) The air in contact with the top-mounted coil is cooled, its volume decreases, and its density increases. The surrounding warmer and less dense air cannot provide sufficient buoyant force to support the colder air, which sinks downward. In the process, warmer air near the bottom is displaced upward and is then cooled by the coil, establishing the convection current. Had the cooling coil been placed at the bottom of the refrigerator, stagnant, cool air would have collected there, with little convection to carry the heat from other parts of the refrigerator to the coil for removal.

Updrafts, or thermals, are caused by the convective movement of air that the ground has warmed . Sometimes, meteorological conditions cause a layer to form in the atmosphere where the temperature increases with increasing altitude. Such a layer is called an inversion layer.

Natural convection, in which a temperature difference causes the density at one place in a fluid to be different from that at another. The forced convection generated by a pump circulates radiator fluid through an automobile engine to remove excess heat.

Conduction Conduction is the process whereby heat is transferred directly through a material, with any bulk motion of the material playing no role in the transfer. Those materials that conduct heat well are called thermal conductors, and those that conduct heat poorly are known as thermal insulators.

The amount of heat Q conducted through the bar from the warmer end to the cooler end depends on a number of factors: Q is proportional to the time t during which conduction takes place (Q t). 2. Q is proportional to the temperature difference T between the ends of the bar (Q T).

3. Q is proportional to the cross-sectional area A of the bar (Q A). 4. Q is inversely proportional to the length L of the bar (Q 1/L). Q    (A DT)t/L.

CONDUCTION OF HEAT THROUGH A MATERIAL The heat Q conducted during a time t through a bar of length L and cross-sectional area A is where T is the temperature difference between the ends of the bar and k is the thermal conductivity of the material. SI Unit of Thermal Conductivity: J/(s·m·C°)

Since the SI unit for thermal conductivity is J·m/(s·m2·C°) or J/(s·m·C°). The SI unit of power is the joule per second (J/s) or watt (W), so the thermal conductivity is also given in units of W/(m·C°). k = QL/(tA DT),

Substance   Thermal Conductivity, k [J/(s·m·C°)]   Metals     Aluminum   240     Brass   110     Copper   390     Iron    79     Lead    35     Silver   420     Steel (stainless)    14 

Gases Air 0.0256 Hydrogen (H2) 0.180 Nitrogen (N2) 0.0258 Oxygen (O2) Other Materials     Asbestos   0.090     Body fat   0.20     Concrete   1.1     Diamond   2450     Glass   0.80     Goose down   0.025     Ice (0 °C)   2.2     Styrofoam   0.010     Water   0.60     Wood (oak)   0.15  Gases     Air   0.0256     Hydrogen (H2)   0.180     Nitrogen (N2)   0.0258     Oxygen (O2)   0.0265 

Styrofoam is an excellent thermal insulator because it contains many small, dead-air spaces. These small spaces inhibit heat transfer by convection currents, and air itself has a very low thermal conductivity.

Example 2. Heat Transfer in the Human Body When excessive heat is produced within the body, it must be transferred to the skin and dispersed if the temperature at the body interior is to be maintained at the normal value of 37.0 °C. One possible mechanism for transfer is conduction through body fat. Suppose that heat travels through 0.030 m of fat in reaching the skin, which has a total surface area of 1.7 m2 and a temperature of 34.0 °C. Find the amount of heat that reaches the skin in half an hour (1800 s).

Example 3. Layered Insulation One wall of a house consists of 0.019-m-thick plywood backed by 0.076-m-thick insulation. The temperature at the inside surface is 25.0 °C, while the temperature at the outside surface is 4.0 °C, both being constant.

The thermal conductivities of the insulation and the plywood are, respectively, 0.030 and 0.080 J/(s·m·C°), and the area of the wall is 35 m2. Find the heat conducted through the wall in one hour (a) with the insulation and (b) without the insulation. (a)

(b)

Conceptual Example 4. An Iced-up Refrigerator

In a refrigerator, heat is removed by a cold refrigerant fluid that circulates within a tubular space embedded within a metal plate. A good refrigerator cools food as quickly as possible. Decide whether the plate should be made from aluminum or stainless steel and whether the arrangement works better or worse when it becomes coated with a layer of ice. The plate should be made from aluminum . When covered with ice, the cooling plate works less well.

Check Your Understanding 1 Two bars are placed between plates whose temperatures are Thot and Tcold (see the drawing). The thermal conductivity of bar 1 is six times that of bar 2 (k1 = 6k2), but bar 1 has only one-third the cross-sectional area ( ). Ignore any heat loss through the sides of the bars. Which statement below correctly describes the heat conducted by the bars in a given amount of time?

c Bar 1 conducts 1/4 the heat as does bar 2; Bar 1 conducts twice the heat as does bar 2;    Bar 1 conducts four times the heat as does bar 2;    Both bars conduct the same amount of heat; c

Radiation Radiation is the process in which energy is transferred by means of electromagnetic waves. In the transfer of energy by radiation, the absorption of electromagnetic waves is just as important as their emission. The surface of an object plays a significant role in determining how much radiant energy the object will absorb or emit.

Since the color black is associated with nearly complete absorption of visible light, the term perfect blackbody or, simply, blackbody is used when referring to an object that absorbs all the electromagnetic waves falling on it.

Since absorption and emission are balanced, a material that is a good absorber, like lampblack, is also a good emitter, and a material that is a poor absorber, like polished silver, is also a poor emitter. THE STEFAN-BOLTZMANN LAW OF RADIATION The radiant energy Q, emitted in a time t by an object that has a Kelvin temperature T, a surface area A, and an emissivity e, is given by where is the Stefan-Boltzmann constant and has a value of 5.67 × 10–8 J/(s·m2·K4).

Example 5. A Supergiant Star The supergiant star Betelgeuse has a surface temperature of about 2900 K (about one-half that of our sun) and emits a radiant power (in joules per second, or watts) of approximately 4 × 1030 W (about 10 000 times as great as that of our sun). Assuming that Betelgeuse is a perfect emitter (emissivity e = 1) and spherical, find its radius.

Example 6. A Wood-Burning Stove A wood-burning stove stands unused in a room where the temperature is 18 °C (291 K). A fire is started inside the stove. Eventually, the temperature of the stove surface reaches a constant 198 °C (471 K), and the room warms to a constant 29 °C (302 K). The stove has an emissivity of 0.900 and a surface area of 3.50 m2. Determine the net radiant power generated by the stove when the stove (a) is unheated and has a temperature equal to room temperature and (b) has a temperature of 198 °C.

(a)

(b)

Check Your Understanding 2 Two identical cubes have the same temperature. One of them, however, is cut in two and the pieces are separated (see the drawing). What is true about the radiant energy emitted in a given time?

a.The cube cut into two pieces emits twice as much radiant energy as does the uncut cube;    b.The cube cut into two pieces emits more radiant energy than does the uncut cube according to    c.The cube cut into two pieces emits the same amount of radiant energy as does the uncut cube;    d.The cube cut into two pieces emits one-half the radiant energy emitted by the uncut cube;    e.The cube cut into two pieces emits less radiant energy than does the uncut cube according to (b)

Applications Insulation inhibits convection between inner and outer walls and minimizes heat transfer by conduction. With respect to conduction, the logic behind home insulation ratings comes directly from Equation The term L/k in the denominator is called the R value of the insulation. Larger R values reduce the heat per unit time flowing through the material and, therefore, mean better insulation.

A thermos bottle minimizes energy transfer due to convection, conduction, and radiation.

In a halogen cooktop, quartz-iodine lamps emit a large amount of electromagnetic energy that is absorbed directly by a pot or pan.

Concepts & Calculations Example 7. Boiling Water Two pots are identical, except that in one case the flat bottom is aluminum and in the other it is copper. Each pot contains the same amount of boiling water and sits on a heating element that has a temperature of 155 °C. In the aluminum-bottom pot, the water boils away completely in 360 s. How long does it take the water in the copper-bottom pot to boil away completely?

Concepts & Calculations Example 8. Freezing Water One half of a kilogram of liquid water at 273 K (0 °C) is placed outside on a day when the temperature is 261 K (–12 °C). Assume that heat is lost from the water only by means of radiation and that the emissivity of the radiating surface is 0.60. How long does it take for the water to freeze into ice at 0 °C when the surface area from which the radiation occurs is (a) 0.035 m2 (as it could be in a cup) and (b) 1.5 m2 (as it could be if the water were spilled out to form a thin sheet)?

(a) Smaller area (b) Larger area

Problem 1 REASONING AND SOLUTION According to Equation 13.1, the heat per second lost is

Problem 8

REASONING To find the total heat conducted, we will apply Equation 13 REASONING To find the total heat conducted, we will apply Equation 13.1 to the steel portion and the iron portion of the rod. In so doing, we use the area of a square for the cross section of the steel. The area of the iron is the area of the circle minus the area of the square. The radius of the circle is one half the length of the diagonal of the square. SOLUTION In preparation for applying Equation 13.1, we need the area of the steel and the area of the iron. For the steel, the area is simply ASteel = L2, where L is the length of a side of the square. For the iron, the area is AIron = R2 – L2. To find the radius R, we use the Pythagorean theorem, which indicates that the length D of the diagonal is related to the length of the sides according to D2 = L2 + L2. Therefore, the radius of the circle is . For the iron, then, the area is

Taking values for the thermal conductivities of steel and iron from Table 13.1 and applying Equation 13.1, we find

PL = e T4A = (0.80)[5.67 *10–8 J/(sm2K4)](307 K)4(1.5 m2) = 604 W Problem 22 REASONING AND SOLUTION a. The radiant power lost by the body is PL = e T4A = (0.80)[5.67 *10–8 J/(sm2K4)](307 K)4(1.5 m2) = 604 W The radiant power gained by the body from the room is Pg = (0.80)[5.67 * 10–8 J/(sm2K4)](298 K)4(1.5 m2) = 537 W The net loss of radiant power is P = PL - Pg = b. The net energy lost by the body is

Problem 28 REASONING AND SOLUTION The rate at which energy is gained through the refrigerator walls is Therefore, the amount of heat per second that must be removed from the unit to keep it cool is

Problem 34

REASONING AND SOLUTION Using Equation 13.1, Before Equation (1) can be applied to the ice-aluminum combination, the temperature T at the interface must be determined. We find the temperature at the interface by noting that the heat conducted through the ice must be equal to the heat conducted through the aluminum: Qice = Qaluminum. Applying Equation 13.1 to this condition, we have

The factors A and t can be eliminated algebraically The factors A and t can be eliminated algebraically. Solving for T gives T = –24.959 °C for the temperature at the interface a. Applying Equation (1) to the ice leads to Since heat is not building up in the materials, the rate of heat transfer per unit area is the same throughout the ice-aluminum combination. Thus, this must be the heat per second per square meter that is conducted through the ice-aluminum combination. b. Applying Equation (1) to the aluminum in the absence of any ice gives: