Physics, 6th Edition
Chapter 18. Transfer of Heat
Chapter 18. Transfer of Heat Note: Refer to Tables 18-1 and 18-2 for thermal conductivities and other constants needed in the solution of the problems in this section.
Thermal Conductivity 18-1. A block of copper has a cross-section of 20 cm2 and a length of 50 cm. The left end is maintained at 00C and the right end is a 1000C. What is the rate of heat flow in watts? k = 385 W/m K; ∆t = 1000C – 00C = 100 C0; 20 cm2 = 20 x 10-4 m2 H=
kA∆t (385 W/m C0 )(20 x 10-4 m 2 )(100 Co ) ; = L 0.50 m
H = 154 W
18-2. In Problem 18-1, what is the heat flow if the copper block is replaced with an block of glass having the same dimensions? kG = 0.80 W/m K; A = 20 x 10-4 m2; L = 0.50 m; ∆t = 100 C0 kB LA k B (6cm ) LB = = = kA kB 2
3cm
;
H = 0.320 W
18-3. A 50-cm long brass rod has a diameter of 3 mm. The temperature of one end is 76 C0 higher than the other end. How much heat is conducted in one minute?
π D 2 π (0.003 m) 2 A= = = 7.07 x 10-6 m 2 ; L = 0.50 m; ∆t = 76 C0 4 4 kA∆t (109 W/m C0 )(7.07 x 10-6 m 2 )(76 Co ) ; H = 0.117 W H= = L 0.50 m LB =
k BLA k B (6 cm) = = kA k B2
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3 cm ;
Q = 7.03 J
Physics, 6th Edition
Chapter 18. Transfer of Heat
18-4. A pane of window glass is 10 in. wide, 16 in. long, and 1/8 in. thick. The inside surface is at 600F and the outside surface is at 200F. How many Btu are transferred to the outside in a time of 2 h? [ L = (1/8) in. = 0.125 in.; A = (10 in.)(16 in.) = 160 in.2 ] 1 ft 2 A = (160 in. ) = 1.11 ft 2 ; 2 144 in. 2
∆t = 600F – 200F = 40 F0
kA∆t τ (5.6 Btu in./ft 2 h F0 )(1.11 ft 2 )(40 F0 )(2 h) ; Q= = L 0.125 in.
Q = 3980 Btu
18-5. One end of an iron rod 30 cm long and 4 cm2 in cross-section is placed in a bath of ice and water. The other end is placed in a steam bath. How many minutes are needed to transfer 1.0 kcal of heat? In what direction does the heat flow?
τ=
QL (1 kcal)(0.30 m) = ; -2 kA∆t (1.2 x 10 kcal/m s C0 )(4 x 10-4 m 2 )(100 C0 )
τ = 625 s;
τ = 10.4 min
The direction of heat flow is from the stem bath to the ice and water. 18-6. A steel plate of thickness 20 mm has a cross-section of 600 cm2. One side is at 1700C and the other is at 1200C. What is the rate of heat transfer? (50.2 W/m K)(600 x 10-4 m 2 )(1700 C - 1200 C) ; H= 0.020 m
H = 7530 W
18-7. How much heat is lost in 12 h through a 3-in. brick firewall 10 ft2 area if one side is at 3300F and the other is at 780F? ∆t = 3300F – 780F = 252 F0; A = 10 ft2; τ = 12 h; L = 3 in. (5 Btu in./ft 2 h F0 )(10 ft 2 )(252 F0 )(12 h) ; Q= 3 in.
250
Q = 50,400 Btu
Physics, 6th Edition
Chapter 18. Transfer of Heat
*18-8. A composite wall 6 m long and 3 m high consists of 12 cm of concrete joined with 10 cm of corkboard. The inside temperature is 100C and the outside temperature is 400C. Find the temperature at the interface of the two materials. [ ∆t = 400C – 100C = 30 C0 ] Hsteel = Hsilver;
LB =
kB L A ( 6 kcB m ) 3 c m= kA 2 kB
50C
t
950C 1 steel
=
(50.2 kcal/m s C0 )(950 C - t ) (406 kcal/m s C0 )(t - 50 C ) ; = 0.30 m 0.60 m
silver
2
t = 22.80C
*18-9. What is the steady-state rate of heat flow through the composite wall of Problem 10-8? H1 = H2 =
(406 kcal/m s C0 )(22.80 C - 50 C)(4 x 10-4 m 2 ) ; 0.60 m
H = 4.83 J/s
Thermal Radiation 18-10. What is the power radiated from a spherical blackbody of surface area 20 cm2 if its temperature is 2500C? R=
P = eσ T 4 ; T = 2500C + 2730 = 523 K; 20 cm2 = 20 x 10-4 m2 A
P = eσ AT 4 = (1)(5.67 x 10-8 W/m 2 K 4 )(20 x 10-4 m 2 )(523 K) 4 ;
P = 8.48 W
18-11. What is the rate of radiation for a spherical blackbody at a temperature of 3270C? Will this rate change if the radius is doubled for the same temperature? T = 3270C + 2730 = 600 K;
e = 1;
R = (1)(5.67 x 10-8 W/m 2 K 4 )(600 K) 4 ; R=
R = eσT4 R = 7.35 kW/m2
P = eσ T 4 so that changing the radius does NOT change the rate of radiation. A
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Physics, 6th Edition
Chapter 18. Transfer of Heat
*18-12. The emissivity of a metallic sphere is 0.3 and at a temperature of 500 K its radiates a power of 800 W. What is the radius of the sphere? R = eσ T 4 = (0.3)(5.67 x 10-8 W/m 2 K 4 )(500 K) 4 = 1063 W/m 2 ;
A = π r2 =
P ; R
r=
P 800 W = ; πR π (1063 W/m 2 )
R=
P A
r = 0.489 m
*18-13. A certain body absorbs 20 percent of the incident thermal radiation, what is its emissivity? What energy will be emitted by this body in 1 min if its surface area is 1 m2 and its temperature is 7270C? e = 20% = 0.2;
T = 2730 + 7270 = 1000 K; τ = 60 s;
R = eσT4 = (0.2)(5.67 x 10-8 W/m2 K4)(1000 K)4; R=
E ; Aτ
R = 11,340 W/m2
E = RAτ = (11,340 W/m 2 )(1 m 2 )(60 s) ;
E = 680 kJ
*18-14. The operating temperature of the filament in a 25-W lamp is 27270C. If the emissivity is 0.3, what is the surface area of the filament? [ T = 27270C + 2730 = 3000 K ] P = eσ AT 4 ;
A=
25 W ; (0.3)(5.67 x 10 W/m 2 K 4 )(3000 K) 4 -8
A = 1.81 x 10-5 m2 or
A = 0.181 cm2
Challenge Problems 18-15. A glass window pane is 60 cm wide 1.8 m tall and 3 mm thick. The inside temperature is 200C and the outside temperature is -100C. How much heat leaves the house through this window in one hour? A = (0.6 m)(1.8 m) = 1.08 m2; ∆t = [20 – (-10)] = 30 C0 Q=
kA∆t τ (0.8 W/m 2 K)(1.08 m 2 )(30 C0 )(3600 s) ; = L 0.003 m 252
Q = 3.11 x 107 J
Physics, 6th Edition
Chapter 18. Transfer of Heat
18-16. A 20-cm thickness of fiberglass insulation covers a 20 m by 15 m attic floor. How many calories of heat are lost to the attic per second if the temperatures on the sides of the insulation are +100C and +240C. [ ∆t = 24 – 10 = 14 C0; A = (20 m)(15 m) = 300 m2 ] Q kA∆t (1 x 10-5 kcal/m s C0 )(300 m 2 )(14 C0 ) ; = = τ L 0.20 m
Q = 0.210 kcal/s
18-17. The bottom of an aluminum pan is 3 mm thick and has a surface area of 120 cm2. How many calories per minute are conducted through the bottom of the pan if the temperature of the outer and inner surfaces are 1140C and 1170C. [ ∆t = 1170 – 1140 = 3 C0 ] Q=
(0.05 kcal/m s C0 )(120 x 10-4 m 2 )(3 C0 ) ; 0.003 m
Q = 36 kcal
18-18. A solid wall of concrete is 80 ft high, 100 ft wide and 6 in. thick. The surface temperatures on the sides of the concrete are 300F and 1000F. What time is required for 400,000 Btu of heat to be transferred? [ A = (80 ft)(100 ft) = 8000 ft2; ∆t = 1000 – 300 = 70 F0 ]
τ=
QL (400, 000 Btu)(6 in.) = ; τ = 0.765 h or 45.9 min kA∆t (5.6 Btu in./ft 2 h F0 )(8000 ft 2 )(70 F0 )
18-19. The bottom of a hot metal pan has an area of 86 cm2 and a temperature of 980C. The pan is placed on top of a corkboard 5 mm thick.. The Formica table top under the corkboard is maintained at a constant temperature of 200C. How much heat is conducted through the cork in 2 min? ∆t = 980 – 200 = 78 C0; A = 86 cm2 = 86 x 10-4 m2; τ = 2 min = 120 s (0.04 W/m C0 )(86 x 10-4 m 2 )(78 C0 )(120 s) ; Q= 0.005 m
253
Q = 644 J
Physics, 6th Edition
Chapter 18. Transfer of Heat
*18-20. What thickness of copper is required to have the same insulating value as 2 in. of corkboard? First find the R-value for 2 in. of corkboard. R = (2 in.)(3.3 ft2 h F0/Btu) = 6.6 ft2 h F0/Btu Copper: L(0.00038 ft2 h F0/Btu) = 6.6 ft2 h F0/Btu; L = 17,370 in. or 1450 ft *18-21. What thickness of concrete is required to have the same insulating value as 6 cm of fiberglass? For the same insulating value, the ratio’s (L/k) must be the same. Lcon L fib = ; kcon k fib
kcon L fib
Lcon =
k fib
=
(0.8 W/m K)(6 cm) ; Lcon = 120 cm (0.04 W/m K)
18-22. A plate-glass window in an office building measures 2 by 6 m and is 1.2 cm thick. It’s outer surface is at 230C and its inner surface is at 250C. How many joules of heat pass through the glass in an hour? A = (2 m)(6 m) = 12 m2; ∆t = 250 – 230 = 2 C0 Q=
(0.8 W/m C0 )(12 m 2 )(2 C0 )(3600 s) ; 0.012 m
Q = 5760 kJ
18-23. What must be the temperature of a black body if its rate of radiation is 860 W/m2? R = eσ T 4 ;
T4 =
R 860 W/m 2 = = 1.517 x 1010 K 4 ; T = 351 K eσ (1)(5.67 x 10-8 W/m 2 K 4 )
18-24. A gray steel ball has an emissivity of 0.75 and when its temperature is 5700C, the power radiated is 800 W. What is the surface area of the ball? [ T = 2730 + 5700 = 843 K ] P = eσ AT 4 ;
A=
P (800 W) = 4 eσ T (0.75)(5.67 x 10-8 W/m 2 K 4 )(843 K) 4
A = 0.0373 m2
or
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373 cm2
Physics, 6th Edition
Chapter 18. Transfer of Heat
18-25. A 25-W lamp has a filament of area 0.212 cm2. If the emissivity is 0.35, what is the operating temperature of the filament? kL B Ak B LB = = = kA kB
(6c m ) 2
T = 4 5.94 x 1013 K 4 ;
3 c m
T = 2776 K
Critical Thinking Questions *18-26. The wall of a freezing plant consists of 6 in. of concrete and 4 in. of corkboard. The temperature of the inside cork surface is -150F and the temperature of the outside surface is 700F. What is the temperature at the interface between the cork and concrete? How much heat is conducted through each square foot 6 in.
ti
4in.
of wall in one hour? -150F
+700F
For steady state: Hcon = Hcork k1 A(700 C - ti ) k2 A[ti - (-150 C)] = L1 L2
Concrete
Cork
(5.6 Btu in./ft 2 h F0 )(700 C - ti ) (0.30 Btu in./ft 2 h F0 )(ti + 150 C ) = 6 in. 4 in. (0.933 Btu/ft2 h)(700C – ti) = (0.075 Btu/ft2 h)(ti + 150F) Solving for ti, we obtain:
ti = 63.70F
The heat conducted per square foot is the same. Suppose we choose the concrete, then H (5.6 Btu in./ft 2 h F0 )(700 C − 63.7 0 C) = ; A (6 in.)
H/A = 5.92 Btu/ft2 h
After some time the heat per unit area per unit of time is the same throughout the wall.
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Physics, 6th Edition
Chapter 18. Transfer of Heat
*18-27. A wooden icebox has walls that are 4 cm thick, and the overall effective surface area is 2 m2. How many grams of ice will be melted in 1 min if the inside temperatures of the walls are 40C and the outside temperatures are 200C?
40C C
Inside and outside temperatures are surface temperatures: kA∆t τ Q= ; L
200 C
A = 2 m2
kA∆t τ Q = mi L f = L
kA∆t τ (0.10 W/m 2 K 4 )(2 m 2 )(60 s)(200 C - 40 C) mi = = ; LL f (0.04 m)(334 x 103 J/kg)
mi = 14.4 g
Note: If the inside and outside temperatures were air temperatures, this would involve both convection and conduction, making the solution much more complicated. *18-28. The filament in a lamp operates at a temperature of 7270C and is surrounded by an envelope at 2270C. If the filament has an emissivity of 0.25 and a surface area of 0.30 cm2, what is the operating power of the lamp? P = eσA(T14 – T24) = eσA[(1000 K)4 – (500 K)4] = eσA(9.375 x 1011 K4) P = (0.25)(5.67 x 10-8 W/m2 K4)(3 x 10-5 m2)(9.375 x 1011 K4) P = 0.399 W *18-29. The thermal conductivity of material A is twice that for material B. Which has the greater R value? When 6 cm of material A is between two walls the rate of conduction is 400 W/m2. What thickness of material B is needed to provide the same rate of heat transfer if other factors remain constant? kA = 2 kB; The R-value is (L/k), so that L L 1 L = = Thus, k A 2k B 2 k B
256
RA = ½RB
Physics, 6th Edition
Chapter 18. Transfer of Heat
If the same rates of heat transfer are to be realized, then the thicknesses must be chosen so that the R values are the same: LA LB = k A kB
LB =
k B LA k B (6 cm) = = 3 cm ; LB = 3.00 cm kA 2k B
Material B needs to be only half the thickness of Material A to provide the same insulating value.
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