ch21.pdf

CHAPTER 21

Thermal Properties and Processes 1* Why does the mercury level first decrease slightly when a thermometer is placed in warm water? The glass bulb warms and expands first, before the mercury warms and expands. 2 A large sheet of metal has a hole cut in the middle of it. When the sheet is heated, the area of the hole will (a) not change. (b) always increase. (c) always decrease. (d) increase if the hole is not in the exact center of the sheet. (e) decrease only if the hole is in the exact center of the sheet. (b) 3 A steel ruler has a length of 30 cm at 20C. What is its length at 100C? Apply Equ. 21-2. DL = (11 10-6)(30)(80) cm = 0.0264 cm;

L = 30.0264 cm 4 A bridge 100 m long is built of steel. If it is built as a single, continuous structure, how much will its length change

from the coldest winter days (30C) to the hottest summer days (40C)? Apply Equ. 21-2. DL = (11 10-6)(100)(70) m = 0.077 m = 7.7 cm 5* (a) Define a coefficient of area expansion. (b) Calculate it for a square and a circle, and show that it is 2

times the coefficient of linear expansion.

(a) g = TA/AD

D. (b) For a square, DA = L2(1 + aDT)2 - L2 = L2(2aDT + a2DT2) = A(2aDT + a2DT2); in the limit

DT 0, DA/A = 2aDT, and g = 2a. For the circle, proceed in same way except that now A = pR2; again, g = 2a. 6 The density of aluminum is 2.70 103 kg/m3 at 0C. What is the density of aluminum at 200C? Apply Equs. 21-4 and 21-5. r = m/V; r = m/(V + DV) = r/(1 + bDT)

b = 3a = 72 10-6 K-1

r = 2.70 103/[1+(72 10-6)(200)] = 2.66 103 kg/m3

Chapter 21 Thermal Properties and Processes

7 A copper collar is to fit tightly about a steel shaft whose diameter is 6.0000 cm at 20C. The inside diameter of the copper collar at that temperature is 5.9800 cm. To what temperature must the copper collar be raised so that it will just slip on the steel shaft, assuming the steel shaft remains at 20C?

aDT = 0.02 cm; use Equ. 21-2 and solve for DT DT = 0.02/(5.98 17 10-6) = 197 Co; T = 217oC 8 Repeat Problem 7 when the temperature of both the steel shaft and copper collar are raised simultaneously. Now RFe = RCu, and both expand Solve for DT and T = (20 + DT)oC

6.0000(1 + 11 10-6DT) = 5.9800(1 + 17 10-6DT)

)10 11 (6.00 )10 17 (5.980.02

66 -- -D = T = 561 Co;

T = 581oC 9* A container is filled to the brim with 1.4 L of mercury at 20C. When the temperature of container and mercury is

raised to 60C, 7.5 mL of mercury spill over the brim of the container. Determine the linear expansion coefficient of the container.

1. Express problem statement in terms of V and DV 2. Apply Equ. 21-4 and solve for bHg - bc 3. Solve for bc and apply Equ. 21-5

VHg = Vc = 1.4 L; DVHg - DVc = 7.5 10-3 L

bHg - bc = [7.5 10-3/(1.4 40)] K-1 = 1.34 10-4 K-1

bc = (1.8 - 1.34) 10-4 K-1 = 0.46 10-4 K-1

a = 15 10-6 K-1

10 A hole is drilled in an aluminum plate with a steel drill bit whose diameter at 20o C is 6.245 cm. In the

process of drilling, the temperature of the drill bit and of the aluminum plate rise to 168C. What is the diameter of the hole in the aluminum plate when it has cooled to room temperature?

1. Find diameter of the hole (steel drill bit) at 168oC 2. Find diameter of the hole in the plate at 20oC

dFe = 6.245(1 + 11 10-6 148) cm = 6.255 cm dAl = 6.255(1 - 24 10-6 148) cm = 6.233 cm

Note that the diameter of the hole in the plate at 20 oC is less than the diameter of the drill bit at 20oC. 11 Len sells trees that double in price when they are over 2.00 m high. To make a standard, he cuts an aluminum rod

2.00 m in length, as measured by a steel measuring tape. That day, the temperature of both the rod and the tape is 25C. What will the tape indicate the length of the rod to be when both the tape and the rod are at (a) 0C and (b) 50C?

1. Apply Equ. 21-2 to the rod and tape 2. Solve for Lrod in terms of Ltape 3. Use numerical values for aFe and aAl. (a) Use negative sign (b) Use positive sign

Lrod = 2.00(1 25aAl) m; Ltape = 2.00(1 25aFe) m

Tape reading = Fe

Al aa

251251

2 m

Tape reading at 0oC = 1.999 m Tape reading at 50oC = 2.001 m

12 A rookie crew was left to put in the final 1 km of rail for a stretch of railroad track. When they finished, the

temperature was 20C, and they headed to town for some refreshments with their coworkers. After an hour or two, one of the old-timers noticed that the temperature had gone up to 25C, so he said, "I hope you left some gaps to allow


for expansion." By the look on their faces, he knew that they had not, and they all rushed back to the work site. The rail had buckled into an isosceles triangle. How high was the buckle?1. In the figure, L is the length at 20oC, L the length at 25oC, and h the height of the buckle.

2. h = 1/2(L2 - L2)1/2

3. L2 = L2(1 + aDT)2 @ L2(1 + 2aDT) 4. h = 1/2L(2aDT)1/2

h = (500)(2 11 10-6 5)1/2 m = 5.24 m

13* A car has a 60-L steel gas tank filled to the top with gasoline when the temperature is 10C. The coefficient of

volume expansion of gasoline is b = 0.900 103 K1. Taking the expansion of the steel tank into account, how much gasoline spills out of the tank when the car is parked in the sun and its temperature rises to 25C?

Spill = DVgas - DVtank = VDT(bgas - btank) Spill = (60)(15)(9 10-4 - 3 11 10-6) L = 0.78 L

14 A thermometer has an ordinary glass bulb and thin glass tube filled with 1 mL of mercury. A temperature change

of 1 C changes the level of mercury in the thin tube by 3.0 mm. Find the inside diameter of the thin glass tube. 1. Net volume change, DV = DVHg - DVglass = ADL, where A is the area of the capillary. 2. DV = V0DT(bHg - bglass); solve for A

3. Find d from 42 /d = A p

Note: 1 mL = 10-6 m3

m1015m103

1027108110 2823

646-

-

--

-. =

) .(A =

-

mm2550m10154 8 . = /.d = p-

15 A mercury thermometer consists of a 0.4-mm capillary tube connected to a glass bulb. The mercury

rises 7.5 cm as the temperature of the thermometer increases from 35C to 43C. Find the volume of the thermometer bulb.

1. See Problem 14. V0 = ADL/[(bHg - bglass)DT] mL 7.7 = m

10531841041057 34

242

0 . ) x (.

= V ---

p

0

16 A grandfather's clock is calibrated at a temperature of 20C. (a) On a hot day, when the temperature is 30C,

does the clock run fast or slow? (b) How much does it gain or lose in a 24-h period? Assume that the pendulum is a thin brass rod of negligible mass with a heavy bob attached to the end.

(a) Tp = 2p(L/g)1/2 is the period of a pendulum; thus as temperature T increases, so does L and Tp, and the clock runs slow. (b) dTp/dT = (dTp/dL)(dL/dT) DTp/Tp = 1/2aDT

h) )(2410 (9.5 = Loss 5-

dTp/dL = 1/2Tp/L; dL/dT = aL.

DTp/Tp = 1/2(19 10-6)(10) = 9.5 10-5

DT = 24 60 60 9.5 10-5 s = 8.21 s lost in 24 h.

17* A steel tube has an outside diameter of 3.000 cm at room temperature (20C). A brass tube has an inside diameter

of 2.997 cm at the same temperature. To what temperature must the ends of the tubes be heated if the


steel tube is to be inserted into the brass tube?

rs(1 + asDT) = rb(1 + abDT);r r

r r = Tssbb

bs

aa --

D

T = T0 + DT

K 125 =K 0003101199721019

9972000366

. .

. . = T

--

D--

T = (20 + 125)oC = 145oC

18 What is the tensile stress in the copper collar of Problem 7 when its temperature returns to 20C? Strain = DL/L; Stress = Y(DL/L) Stress = (11 10-10)(0.02/5.98) N/m2 = 3.68 108 N/m2 19 Mountaineers say that you cannot hard boil an egg on the top of Mount Rainier. This is true because (a) the air is too cold to boil water. (b) the air pressure is too low for stoves to burn. (c) boiling water is not hot enough to hard boil the egg. (d) the oxygen content of the air is too low. (e) eggs always break in their backpacks.

(c) (Actually, it can be hard boiled, but it does take quite a bit longer than at sea level.) 20 Which gases in Figure 21-6 cannot be liquefied by applying pressure at 20C? Gases for which Tc < 293 K. These are He, A, Ne, H2, O2, NO. 21* The phase diagram in Figure 21-14 can be interpreted to yield information on how the boiling and melting

points of water change with altitude. (a) Explain how this information can be obtained. (b) How might this information affect cooking procedures in the mountains? (a) With increasing altitude P decreases; from curve OF, T of liquid-gas interphase diminishes, so the boiling temperature decreases. Likewise, from curve OH, the melting temperature increases with increasing altitude. (b) Boiling at a lower temperature means that the cooking time will have to be increased. 22 For the phase diagram given in Figure 21-14, state what changes (if any) occur for each line segmentAB, BC,

CD, and DEin (a) volume and (b) phase. (c) For what type of substance would OH be replaced by OG? (d) What is the significance of point F?

(a) and (b). AB: solid sublimates to vapor; volume increases. BC: vapor condenses to liquid; volume decreases. CD: liquid freezes; volume increases. DE: solid changes to liquid; volume decreases.

(c) For most materials, the density increases on solidification; for these materials, the phase diagram would have the shape OG. (d) F is the critical point. 23 (a) Calculate the volume of 1 mol of steam at 100C and a pressure of 1 atm, assuming that it is an ideal gas.

(b) Find the temperature at which the steam will occupy the volume found in part (a) if it obeys the van der Waals equation with a = 0.55 Pam6 / mol2 and b = 30 cm3/mol.

(a) V = nRT/P (b) Use Equ. 21-6; substitute numerical values

V = 1 8.314 373/101.3 103 m3 = 0.0306 m3 = 30.6 L


3148

)103010(30.6)10(30.6

55010011 6323

5

.

. + .

= T

---

-

T = 374 K

24 From Figure 21-4, find (a) the temperature at which water boils on a mountain where the atmospheric pressure is

70 kPa, (b) the temperature at which water will boil in a container in which the pressure has been reduced to 0.5 atm, and (c) the pressure at which water will boil at 115C.

(a) At 70 kPa, the boiling point is at T = 90oC; (b) at 0.5 atm, Tboil = 82 oC; (c) for Tboil = 115oC, P = 170 kPa. 25* The van der Waals constants for helium are a = 0.03412 L2 atm / mol2 and b = 0.0237 L/mol. Use these data to

find the volume in cubic centimeters occupied by one helium atom and to estimate the radius of the atom. In Equ. 21-6, b = volume of 1 mol of molecules For He, 1 molecule = 1 atom V = (4/3)pr3; solve for r

(0.0237 L/mol)(1 mol/6.022 1023 atoms)(103 cm3/1 L) = 3.94 10-23 cm3/atom. r = (3 3.94 10-23/4p)1/3 = 2.11 10-8 cm = 0.211 nm

26 (a) For a van der Waals gas, show that the critical temperature is 8a/27Rb and the critical pressure is a/27b2.

(b) Rewrite the van der Waals equation of state in terms of the reduced variable Vr = V/Vc , Pr = P/Pc , and Tr = T/Tc. (a) At the critical point, dP/dV = 0 and d2P/dV2 = 0. From Equ. 21-6, dP/dV = -nRT/(V - nb)2 + 2an2/V3 = 0 (1) and d2P/dV2 = 2nRT/(V - nb)3 - 6an2/V4 = 0 (2). From (1), 2an2/V3 = nRT/(V - nb)2 (1a); from (2), 6an2/V4 = 2nRT/(V - nb)3 (2a). Dividing (1a) by (2a) gives 1/2(V - nb) = V/3; Vc = 3nb (3). Now substitute Vc from (3) into (1a): RT/4nb2 = 2a/27nb3; Tc = 8a/27Rb. Now substitute Tc and Vc into Equ. 21 6: Pc = a/27b2. (b) Using the results from (a) in Equ. 21-6 one obtains (Pr + 3/Vr3)(3Vr - 1) = 8Tr. 27 A copper bar 2 m long has a circular cross section of radius 1 cm. One end is kept at 100C and the other end is

kept at 0C. The surface of the bar is insulated so that there is negligible heat loss through it. Find (a) the thermal resistance of the bar, (b) the thermal current I, (c) the temperature gradient DT /Dx, and (d) the temperature of the bar 25 cm from the hot end.

(a) R = Dx/kA (b) I = DT/R (c) Substitute numerical values (d) T = T0 + (dT/dx)Dx

R = (2 m)/[(401W/m.K)(p 10-4 m2)] = 15.9 K/W I = 100/15.9 W = 6.3 W

DT/Dx = 100/2 K/m = 50 K/m T = 0oC + 1.75 50oC = 87.5oC

28 A 20 30-ft slab of insulation has an R factor of 11. How much heat (in Btu per hour) is conducted through the

slab if the temperature on one side is 68F and that on the other side is 30F? I = DT/R = ADT/Rf I = (38)(600)/11 Btu/h = 2073 Btu/h 29* Two metal cubes with 3-cm edges, one copper (Cu) and one aluminum (Al), are arranged as shown in Figure 21-

15. Find (a) the thermal resistance of each cube, (b) the thermal resistance of the two-cube system, (c) the thermal current I, and (d) the temperature at the interface of the two cubes.


(a) Use Equ. 21-10; substitute numerical values (b) R = RCu + RAl (c) I = DT/R (d) ICu = IAl = I; DTCu = ICuRCu

RCu = 1/(0.03 401) = 0.0831 K/W; RAl = 0.141 K/W R = 0.224 K/W I = 80/0.224 W = 358 W

DTCu = 358 0.0831 K = 29.7 K; T = 100 - 29.7 = 70.3oC

30 The cubes in Problem 29 are rearranged in parallel as shown in Figure 21-16. Find (a) the thermal current

carried by each cube from one side to the other, (b) the total thermal current, and (c) the equivalent thermal resistance of the two-cube system.

(a) Apply Equ. 21-9 to each cube (b) I = ICu + IAl (c) Requ = DT/I

ICu = 80/0.0831 W = 963 W; IAl = 80/0.141 W = 567 W I = 1530 W R = 80/1530 K/W = 0.0523 K/W

31 A spherical shell of thermal conductivity k has inside radius r1 and outside radius r2 (Figure 21-17). The inside of

the shell is held at a temperature T1, and the outside at temperature T2. In this problem, you are to show that the thermal current through the shell is given by

)T T ( r rrkr

= I 1212

214 --

p

Consider a spherical element of the shell of radius r and thickness dr. (a) Why must the thermal current through each such element be the same? (b) Write the thermal current I through such a shell element in terms of the area A = 4pr2, the thickness dr, and the temperature difference dT across the element. (c) Solve for dT in terms of dr and integrate from r = r1 to r = r2. (d) Show that when r1 and r2 are much larger than r2- r1, Equation 21-22 is the same as Equation 21-7.

(a) From conservation of energy, the thermal current through each shell must be the same. (b) I = -kA(dT/dr) = -4pkr2(dT/dr); note the minus sign - the heat current is directed opposite to temperature gradient.

(c) dT = - (I/4pk)(dr/r2); )T T(r rrkr = I ;

r

rkI

=T T ;r

drk

I = dT

r

r

T

T

2

1

1212

21

21122

41144

2

1

--

--- p

pp

(d) For r2 - r1


33* For a boiler at a power station, heat must be transferred to boiling water at the rate of 3 GW. The boiling water passes through copper pipes having a wall thickness of 4.0 mm and a surface area of 0.12 m2 per meter length of pipe. Find the total length of pipe (actually there are many pipes in parallel) that must pass through the furnace if the steam temperature is 225C and the external temperature of the pipes is 600C.

1. From Equ. 21-7, A = IDx/kDT 2. L = A/(0.12 m)

A = (3 109)(4 10-3)/[(401)(375)] m2 = 79.8 m2 L = 665 m

34 A steam pipe of length L is insulated with a layer of material of thermal conductivity k . Find the rate of heat

transfer if the temperature outside the insulation is t1 , the temperature inside is t2, the outside radius of the insulation is r1, and the inside radius is r2.

Proceed as in Problem 31. Consider an element with a cylindrical area of length L, radius r, and thickness dr. The heat current is I = -2pkLr(dT/dr). Thus, dT = -[I/(2pkL)]dr/r. Integrate from T1 to T2 and from r1 to r2 and solve for the heat current I. I = 2pkL(T1 - T2)/ln(r1/r2).

Note: If we use the above result in Problem 33 (take 0.12 m2 to be the outside area per meter of pipe) then r1 = 1.91 cm and r2 = 1.51 cm. Solving for L one obtains L = 746 m. 35 Brine at -16C circulating through copper pipes with walls 1.5 mm thick is used to keep a cold room at 0C. The

diameter of each pipe is very large compared to the thickness of its walls. By what fraction is the transfer of heat reduced when the pipes are coated with a 5-mm layer of ice?

1. Use Rtot = RCu + Rice; R = Dx/kA. Also I = DT/R 2. Substitute numerical values.

Itot/ICu = RCu/R = [1 + (DxicekCu)/(DxCuk ice)]-1 Itot/ICu = 4.43 10-4; I reduced by a factor of 2260.

36 If the absolute temperature of an object is tripled, the rate at which it radiates thermal energy (a) triples. (b) increases by a factor of 9. (c) increases by a factor of 27. (d) increases by a factor of 81. (e) depends on whether the absolute temperature is above or below zero. (d) The energy radiated is proportional to T4. 37* Calculate lmax for a human blackbody radiator, assuming the surface temperature of the skin to be 33C. Use Equ. 21-21 and substitute numerical values. lmax = (2.898 10-3)/(273 + 33) m = 9.47 mm 38 The heating wires of a 1-kW electric heater are red hot at a temperature of 900C. Assuming that 100% of the

heat output is due to radiation and that the wires act as blackbody radiators, what is the effective area of the radiating surface? (Assume a room temperature of 20C.)

From Equ. 21-20, A = Pnet/[es(T4 - T04)] A = 103/[1 5.67 10-8(11734 - 2934)] m2 = 9.35 10-3 m2

39 A blackened, solid copper sphere of radius 4.0 cm hangs in a vacuum in an enclosure whose walls have a

temperature of 20C. If the sphere is initially at 0C, find the rate at which its temperature changes, assuming that heat is transferred by radiation only.


1. dQ/dt = mc(dT/dt) = Pnet. Find Pnet

2. Solve for dT/dt; m = (4/3)prr3, c = 0.386 kJ/kg.K

Pnet = -4p 16 10-4 5.67 10-8(2934 - 2734) W = -2.07 W dT/dt = -2.07 3/(4p8.961036410-6386) K/s = -2.23 10-3 K/s

40 The surface temperature of the filament of an incandescent lamp is 1300C. If the electric power input is doubled,

what will the temperature become? Hint: Show that you can neglect the temperature of the surroundings.

1. Pnet = esA(T4 - T04) = esAT4[1 - (T0/T)4] 2. T P1/4

(273/1573)4 = 9 10-4


(c) 46 Which heat-transfer mechanisms are most important in the warming effect of a fire in a fireplace? Radiation and convection. 47 Which heat-transfer mechanism is important in the transfer of energy from the sun to the earth? Radiation is the only mechanism. 48 Two cylinders made of materials A and B have the same lengths; their diameters are related by dA = 2dB. When

the same temperature difference is maintained between the ends of the cylinders they conduct heat at the same rate. Their thermal conductivities are related by

(a) kA = kB/4 (b) kA = kB/2 (c) kA = kB (d) kA = 2kB (e) kA = 4kB (e) From Equ. 21-7, kd2 must be constant. 49* A steel tape is placed around the earth at the equator when the temperature is 0C. What will the clearance

between the tape and the ground (assumed to be uniform) be if the temperature of the tape rises to 30C? Neglect the expansion of the earth.

From Equ. 21-2, DR = RaDT DR = 6.38 106 11 10-6 30 m = 2.1 103 m = 2.1 km

50 Use the result of Problem 31 (Equation 21-22) to calculate the wall thickness of the hemispherical igloo of Problem

32 without assuming that the inner surface area equals the outer surface area. Set r2 = r1 + Dr; from Problem 32 (igloo is a

hemisphere) Tkr I

Tkr = rD-

DD

1

21

22

pp

For r1 = 2 m, DT = 40 K, I = 440 W, k = 0.209 W/m.K, one obtains Dr = 0.63 m = 63 cm.

51 Show that change in the density of an isotropic material due to an increase in temperature DT is given by

T.- = DD brr

dT = d ; = V VM

= dTdV

dV d

= dT d

; VM

= 2

brrrbbrr

r --- ; Dr = - brDT

52 The solar constant is the power received from the sun per unit area perpendicular to the sun's rays at the mean distance of the earth from the sun. Its value at the upper atmosphere of the earth is about 1.35 kW/m2. Calculate the effective temperature of the sun if it radiates like a blackbody. (The radius of the sun is 6.96 108 m.) 1. Determine the total power radiated. R = 1.5 1011 m is the sunearth distance. 2. Use Equ. 21-17; A = 4pRS2, where RS = 6.96 108 m

P = (1.35 kW/m2)(4psR2) = (1.35 kW/m2)4p (1.5 1011 m)2 = 38.2 1022 kW = 3.82 1026 W T = [3.82 1026/(5.67 10-8 4p 4.84 1017)]1/4 = 5769 K

53* Lou has patented a cooking timer, which he is marketing as "Nature's Way: Taking You Back To Simpler Times."


The timer consists of a 28-cm copper rod having a 5.0-cm diameter. Just as the lower end is placed in boiling water, an ice cube is placed on the top of the rod. When the ice melts completely, the cooking time is up. A special ice cube tray makes cubes of various sizes to correspond to the boiling time required. What is the cooking time when a 30-g ice cube at 5.0C is used?

1. Find t1, time to raise the ice cube from -5oC to 0oC. t1 = DQ/[kA(DT/Dx)]; DQ = mc (5 K)

2. t2, time to melt ice = mLDx/kADT 3. The total time = t1 + t2

DQ = (0.03 kg)(2.05 103 J/kg.K)(5 K) = 307.5 J t1 = 307.5/[401 (p 25 10-4/4) (102.5/0.28)] s = 1.07 s

t2 = s 35.6 = s100)41025(401280105333030

4

3

/

...

- p

ttot = 36.7 s 54 To determine the R value of insulating material that comes in sheets of 2

1 s-in thickness, you construct a cubical

box of 12 in per side and place a thermometer and a 100-W heater inside the box. After thermal equilibrium has been attained, the temperature inside the box is 90C when the external temperature is 20C. Determine the R value of this material.

1. I = kA(DT/Dx); Rf = Dx/k = ADT/I 2. Convert to U.S. customary units

Rf = 6(0.3048 m)2(70 K)/(100 W) = 0.39 K.m2/W

Rf = 2.2 h.ft2.Fo/Btu 55 A 2-cm-thick copper sheet is pressed against a sheet of aluminum. What should be the thickness of the aluminum

sheet so that the temperature of the copperaluminum interface is (T1 + T2)/2, where T1 and T2 are the temperatures at the copperair and aluminumair interfaces?

DTCu = DTAl = 1/2(T1 - T2) = (I/A)(tCu/kCu) = (I/A)(tAl/kAl)

tAl = tCu(kAl/kCu) = (2 cm)(237/401) = 1.18 cm

56 At a temperature of 20C, a steel bar of radius 2.2 cm and length 60 cm is jammed horizontally perpendicular

between two vertical concrete walls. With a blowtorch, the temperature of the bar is raised to 60C. Find the force exerted by the bar on each wall.

DL = LaDT; F = AYDL/L = AYaDT F = p(2.2 10-2)2(2.0 1011)(11 10-6)(40) N = 1.34 105 N

57* (a) From the definition of b, the coefficient of volume expansion (at constant pressure), show that b = 1/T for an

ideal gas. (b) The experimentally determined value of b for N2 gas at 0C is 0.003673 K1. Compare this value with the theoretical value b = 1/T, assuming that N2 is an ideal gas.

(a) For an ideal gas, V = nRT/P; b = (1/V)(dV/dT) = (P/nRT)(nR/P) = 1/T. (b) 1/273 = 0.003663 is within 0.3 % of the experimental value. 58 One way to construct a device with two points whose separation remains the same in spite of temperature

changes is to bolt together one end of two rods having different coefficients of linear expansion as in the arrangement shown in Figure 21-18. (a) Show that the distance L will not change with temperature if the lengths LA and LB are


chosen such that LA/LB = aB/aA. (b) If material B is steel, material A is brass, and LA = 250 cm at 0C, what is the value of L?

(a) We want L = (LB - LA) = (LB + aBLBDT) - (LA + aALADT). Therefore, aBLB = aALA or LA/LB = aB/aA. (b) From (a) LB = LSteel = (250 cm)(19/11) = 432 cm, and L = 182 cm. 59 On the average, the temperature of the earth's crust increases 1.0 C for every 30 m of depth. The average

thermal conductivity of the earth's crust is 0.74 J/msK. What is the heat loss of the earth per second due to conduction from the core? How does this heat loss compare with the average power received from the sun? (The solar constant is about 1.35 kW/m2.)

1. Heat current/m2 = I/m2 = k(DT/Dx) 2. dQ/dt = kA(DT/Dx)

I/m2 = 0.74/30 W/m2 = 0.0247 W/m2 < 0.002% of the solar constant. dQ/dt = 4p(6.38 106)2 0.0247 W = 1.26 1010 kW

60 A copper-bottomed saucepan containing 0.8 L of boiling water boils dry in 10 min. Assuming that all the heat flows

through the flat copper bottom, which has a diameter of 15 cm and a thickness of 3.0 mm, calculate the temperature of the outside of the copper bottom while some water is still in the pan.

1. I = DQ/Dt = kA(DT/Dx); DQ = mLv; solve for DT 2. m = 0.8 kg, Lv = 2257 kJ/kg, A = .0225p/4 m2,

Dx = 3 10-3 m, k = 401 W/m.K.

DT = mLvDx/kADt

DT = 40.82257103310-3/4010.0225p600 = 1.3 K; Tout = 100oC + DT = 101.3oC.

61* A hot-water tank of cylindrical shape has an inside diameter of 0.55 m and inside height of 1.2 m. The tank is

enclosed with a 5-cm-thick insulating layer of glass wool whose thermal conductivity is 0.035 W/mK. The metallic interior and exterior walls of the container have thermal conductivities that are much greater than that of the glass wool. How much power must be supplied to this tank to maintain the water temperature at 75C when the external temperature is 1C?

We will do this problem twice. First, we shall disregard the fact that the surrounding insulation is cylindrical. We shall then repeat the problem, using the result of Problem 34.

(a) 1. Find the total area 2. Use Equ. 21-7 (b) 1. Find I through top and bottom surfaces: I1 2. Find Ic through cylindrical surface (see Problem 34) 3. Find the total heat loss I = I1 + Ic

Atot = [2 (p/4)(0.55)2 + p 0.55 1.2] m2 = 2.55 m2 I = (0.035)(2.55)(74/0.05) W = 132 W I1 = (0.035)[(p/2)(0.55)2](74/0.05) W = 24.6 W Ic = 2p(0.035)(1.2)(74)/ln(0.65/0.55) W = 97.4 W I = (24.6 + 97.4) W = 122 W

62 The diameter of a rod is given by d = do(1 + ax), where a is a constant and x is the distance from one end. If the

thermal conductivity of the material is k what is the thermal resistance of the rod if its length is L?

I = -kA(dT/dx); ) + ax(d)/A = (22

0 14p ; aL)(kdpI L

= TT ; + ax

dx

kd

IdT = -

LT

T1

4)1(

420

1220

20

2

1

- p

R = +aL)(kd

4L =

IT

120pD


63 A solid disk of radius R and mass M is spinning in a frictionless environment with angular velocity w1 at

temperature T1. The temperature of the disk is then changed to T2. Express the angular velocity w2, rotational kinetic energy E2, and angular momentum L2 in terms of their values at the temperature T1 and the linear expansion coefficient a of the disk.

Let DT = T2 - T1. 1. No torque: L2 = L1. 2. I2 = MR22 = MR12(1 + aDT)2 = I1(1 + 2aDT + a2DT2) = I1(1 + 2aDT), neglecting the higher order term. I1w1 = I2w2, w2 = w1(I1/I2) = w1(1 - 2aDT). E2 = L22/2I2 = L12/2I2; E1 = L12/2I1; E2 = E1(I1/I2) = E1(1 - 2aDT). 64 A small pond has a layer of ice 1 cm thick floating on its surface. (a) If the air temperature is 10C, find the rate

in centimeters per hour at which ice is added to the bottom of the layer. The density of ice is 0.917 g/cm3. (b) How long does it take for a 20-cm layer to be built up?

(a) To freeze m kg of water, Q = mLf. To determine the rate of freezing, dQ/dt = Lf(dm/dt) = LfrA(dx/dt). But we also have dQ/dt = kADT/x, where x is thickness of the ice. dx/dt = (k /Lfr)DT/x

(b) t LTk

) =xx( dt; LTk

=x dxf

if

t

f

x

x

f

irrD

-D

220

21

dx/dt = (0.592/333.5 103 917)(10/0.01) m/s = 1.94 mm/s = 6.97 mm/h t = [(333.5 103 917)/(2 0.592 10)](0.04 - 10-4) s = 1.03 106 s @ 12 days

65* A body initially at a temperature Ti cools by convection and radiation in a room where the temperature is T0. The

body obeys Newton's law of cooling, which can be written ),T hA(T = dQ/dt 0- where A is the area of the body and

h is a constant called the surface coefficient of heat transfer. Show that the temperature T at any time t is given by

,e)T 0i hAt / mc0 T( + T = T -- where m is the mass of the body and c is its specific heat.

1. dQ = -mcdT is heat loss as T diminishes by dT. Thus, dT = - (1/mc)dQ and dT/dt = - (hA/mc)(T - T0)

2. t mchA

- = T -TTT ; dt

mchA

- = TT -

dT

i

tT

T i

-

0

0

00

ln , where Ti is the initial temperature.

Take the antilog and solve for T to obtain T = T0 + (Ti - T0)e-hAt/mc. 66 Two 200-g copper containers, each holding 0.7 L of water, are connected by a 10-cm copper rod of cross-

sectional area 1.5 cm2. Initially, one container is at 60C; the second is maintained at 0C. (a) Show that the temperature tc of the first container changes over time t according to

et t / RC- t = 0cc

where tc0 is the initial temperature of the first container, R is the thermal resistance of the rod, and C is the total heat capacity of the container plus the water. (b) Evaluate R, C, and the "time constant" RC. (c) Show that the total amount of heat Q conducted in time t is

)1(0c e Ct = Q t / RC--


(d) Find the time it takes for the temperature of the first container to be reduced to 30C. (a) Heat loss dQ = -(mccc + mwcw)dTc, where mc and cc are the mass and specific heat of the container and mw and cw are the mass and specific heat of the water. Let the sum of those heat capacities be C. Then we can write dQ/dt = -C(dTc/dt). But we also have, from Equ. 21-7, dQ/dt = kA(Tc - 0)/Dx = Tc/R. We therefore have

.eT = T t ;RC

= TT dt;

RC =

TdT t/RC

ccc

ct

c

cT

T

c

c

--

- 0

000

1ln

1

(b) R = Dx/kA; R = 0.1/(401 1.5 10-4) K/W = 1.66 K/W; C = (mccc + mwcw) C = (0.2 386 + 0.7 4180) J/K = 3 kJ/K; RC = 4985 s = 1.38 h (c) To determine Q we integrate dQ = (Tc/R)dt, where Tc = Tc0e-t/RC.

)e( ; Q = C Tdte R

TdQ = t/RCc

-t/RCt

cQ

-- 100

0

0

(d) Tc = 1/2Tc0; exp(-t/RC) = 1/2; t = RC ln (2) t = 1.38 ln(2) = 0.96 h 67 Liquid helium is stored in containers fitted with 7-cm-thick "superinsulation" consisting of a large number of layers

of very thin aluminized Mylar sheets. The rate of evaporation of liquid in a 200-L container is about 0.7 L per day. Assume the container is spherical and that the external temperature is 20C. The specific gravity of liquid helium is 0.125 and the latent heat of vaporization is 21 kJ/kg. Estimate the thermal conductivity of superinsulation.

1. Find rate of loss in kg/s. dm/dt = r(dV/dt)

2. I = Lv(dm/dt) = kA(DT/Dx); k = T A(dm/dt) x LvD

D

A = 4p(3V/4p)2/3

dm/dt = (0.125 103 kg/m3)(0.7 10-3 m3)/(86400 s) = 1.01 10-6 kg/s A = 4p(3 2.0x10-1/4p)2/3 = 1.65 m2

k = (21103710-21.0110-6)/(1.65288) W/m.K

= 3.1 10-6 W/m.K

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