Heat Transfer
EXERCISES, Q41 - Q50
41. A solid aluminum sphere and a solid copper sphere of twice the radius are heated to the same temperature and are allowed to cool under identical surrounding temperatures. Assume that the emissivity of both the spheres is the same. Find the ratio of (a) the rate of heat loss from the aluminum sphere to the rate of heat loss from the copper sphere and (b) the rate of fall of temperature of the aluminum sphere to the rate of fall of the temperature of the copper sphere. The specific heat capacity of aluminum = 900 J/kg-°C and that of copper = 390 J/kg -°C. The density of copper = 3.4 times the density of aluminum.
Answer: Let the radius of the aluminum sphere = r and that of the copper sphere = 2r.
The surface area of the aluminum sphere, A =4Οr².
The surface area of the copper sphere, A' = 4Ο(2r)² =16Οr².
If the emissivity and the temperature of both the spheres are e and T then the heat radiation rate from the aluminum sphere,
u = eπAT,
(Where π is the Stefen constant)
= eπ(4Οr²)T
And the heat radiation from the copper sphere,
u' = eπA'T
=eπ(16Οr²)T
(a) Hence the ratio of the rate of heat loss from the aluminum sphere to the rate of heat loss from the copper sphere
=u/u'
= eπ(4Οr²)T/eπ(16Οr²)T
= 1:4.
(b) Suppose the rate of the temperature fall from the aluminum sphere =dT/dt and from the copper sphere =dT'/dt, then the rate of heat radiation from the aluminum sphere,
u = msdT/dt,
and from the copper sphere,
u' = m's'dT'/dt
Where m and m' are masses and s and s' are specific heat capacities of aluminum and copper spheres respectively. If d is the density of aluminum then density of copper =3.4d. Now,
u/u'=
{(4/3)Οr³d*900*dT/dt}/{(4/3)Ο8r³*3.4d*390*dT'/dt}
=(900*dT/dt)/(8*3.4*390*dT'/dt)
=(1/11.79)(dT/dt)/(dT'/dt)
But u/u' = 1/4, so
(dT/dt)/(dT'/dt) =11.79/4
=2.9:1.
42. A 100 W bulb has tungsten filaments of total length 1.0 m and radius 4x10⁻⁵ m. The emissivity of the filament is 0.8 and π = 6.0x10⁻⁸ W/m²-K⁴. Calculate the temperature of the filament when the bulb is operating at correct wattage.
Answer: Heat radiation per unit time,
u = 100 W, e = 0.8,
Area of the wire, A =2Ο(4x10⁻⁵)*1.0 m², Let the temperature of the wire = T.
Now, u =eπAT⁴
→T⁴ = {100/0.8*6.0x10⁻⁸*2Ο(4x10⁻⁵)}
→T = (0.83x10¹³)1/4
= 1697.3 K
≈ 1700 K.
43. A spherical ball of surface area 20 cm² absorbs any radiation that falls on it. It is suspended in a closed box maintained at 57°C. (a) Find the amount of radiation falling on the ball per second, (b) Find the net rate of heat flow to or from the ball at an instant when its temperature is 200°C. Stefen constant =6.0x10⁻⁸ W/m²-K⁴.
Answer: (a) Since the ball absorbs any radiation falling on it, its emissivity, e =1. Hence the radiation absorbed per second,
u =πAT⁴.
Here, T =273+57 =330 K,
A = 20 cm² =20x10⁻⁴ m² =2x10⁻³ m²
π = 6x10⁻⁸ W/m²-K⁴
So, u=6x10⁻⁸*2x10⁻³*330⁴ J/s
=1.42 J/s
(b) The temperature of the ball T'=200°C =273+200 =473 K.
The ball is at a higher temperature, hence the net flow of heat will be from ball to the surrounding.
The net rate of heat flow from the ball,
=πA(T'⁴-T⁴)
=6x10⁻⁸*2x10⁻³*(473⁴-330⁴)
=4.58 W.
44. A spherical tungsten piece of radius 1.0 cm is suspended in an evacuated chamber maintained at 300 K. The piece is maintained at 1000 k by heating it electrically. Find the rate at which the electrical energy must be supplied. The emissivity of tungsten is 0.30 and the Stefen constant π is 6.0x10⁻⁸ W/m²-K⁴.
Answer: The rate of electrical energy to be supplied will be equal to the net loss rate of heat energy which will be
u =eπA(T'⁴-T⁴)
Here, e =0.3,
A=4Ο*1² cm² =4Ο*10⁻⁴ m²
T' =1000 K, T =300 K
Now,
u =0.3*6x10⁻⁸*4Ο*10⁻⁴(1000⁴-300⁴)
=22.4 W.
45. A cubical block of mass 1.0 kg and edge 5.0 cm is heated to 227°C. It is kept in an evacuated chamber maintained at 27°C. Assuming that the block emits radiation like a blackbody, find the rate at which the temperature of the block will decrease. The specific heat capacity of the material of the block is 400 J/kg-K.
Answer: Mass of the block, m = 1 kg.
The surface area of the block, A = 6*5² cm² =150 cm² =0.015 m².
The temperature of the block,
T' = 227°C = 273+227 =500 K.
The temperature of the chamber,
T =27°C =273+27 =300 K.
Emissivity, e = 1.
Rate of heat radiation,
u =πA(T'⁴-T⁴)
But u =ms(dT/dt)
→dT/dt =πA(T'⁴-T⁴)/ms
=6x10⁻⁸*0.015*(500⁴-300⁴)/(1*400)
=0.12 K/s or 0.12°C/s.
46. A copper sphere is suspended in an elevated chamber maintained at 300 K. The sphere is maintained at a constant temperature of 500 K by heating it electrically. A total of 210 W of electric power is needed to do it. When the surface of the copper sphere is completely blackened, 700 W is needed to maintain the temperature of the sphere. Calculate the emissivity of copper.
Answer: Let the emissivity of copper = e. The power needed to maintain the temperature of the sphere is equal to the net radiation per second by the sphere. So,
u =eπA(T'⁴-T⁴)
When the copper surface is blackened,
e =1, Now the net radiation per second,
u' =πA(T'⁴-T⁴)
The ratio of the two,
u/u' = e,
→e =210/700 = 0.30.
47. A spherical ball A of surface area 20 cm² is kept at the center of a hollow spherical shell B of area 80 cm². The surface of A and the inner surface of B emit as black bodies. Assume that the thermal conductivity of the material of B is very poor and that of A is very high and that the air between A and B has been pumped out. The heat capacities of A and B are 42 J/°C and 82 J/°C respectively. Initially, the temperature of A is 100°C and that of B is 20°C. Find the rate of change of temperature of A and that of B at this instant. Explain the effects of the assumptions listed in the problem.
Answer: Temperature of A, T' =100°C =273+100 =373 K, Area, A' =20 cm² =0.002 m².
Temperature of B, T = 20°C =273+20 =293 K. Area, A =80 cm² =0.008 m².
Emissivity e for both is =1.
Since all the heat radiating out from A falls on the inner surface of B, all of this will be absorbed by B. Heat radiated out by A and received by B = πA'T'⁴
Heat radiated out by B =πAT⁴. But only a fraction of it (A'/A) will fall on A and the rest will fall on the inner surface of B itself and absorbed by it.
Hence net heat radiation received by B,
u =πA'T'⁴ -πAT⁴*(A'/A)
=6x10⁻⁸*0.002(373⁴-293⁴)
=1.44 J/s
But it will be equal to (ms)dT/dt
Hence, (ms)dT/dt =1.44
→dT/dt =1.44/82 = 0.01°C/s
So the rate of change of temperature of B =0.01°C/s.
Now let us consider the surface of A. Since there is no air between the balls, the surrounding temperature can not be taken as T =293 K. It is receiving heat radiation from B at the rate of (A'/A)πAT⁴ =πA'T⁴. From the given assumption conductivity of A is very high which means whatever heat radiation it is receiving it gets immediately spread evenly to whole of the sphere A and rise in its temperature will be instantaneous. So the rise rate of the temperature due to this received radiation from B,
=πA'T⁴/(m's')
=6x10⁻⁸*0.002*293⁴/(42)
=0.02 °C/s.
This is increase of temperature is in a second which is very small in comparison to the temperature of A =373 K. Hence the instantaneous rise in temperature of A due to the received radiation from B is negligible and the heat radiating out from A will be,
u' =πA'T'⁴
=6x10⁻⁸*0.002*373⁴
= 2.32 J/s
Hence the rate of change of temperature of A =u'/(m's')
=2.32/42
=0.05°C/s.
The effects of assumptions in the problem are obvious. If the conductivity of sphere A was not very high, the net radiation received by A would be,
=πA'(T'⁴-T⁴)
=6x10⁻⁸*0.002(373⁴-293⁴)
=1.44 J/s
And the change in the temperature would have been =1.44/42 =0.03°C/s.
48. A cylindrical rod of length 50 cm and cross-sectional area 1 cm² is fitted between a large ice chamber at 0°C and an evacuated chamber maintained at 27°C as shown in figure (28-E12). Only small portions of the rod are inside the chambers and the rest is thermally insulated from the surrounding. the cross-section going into the evacuated chamber is blackened so that it completely absorbs any radiation falling on it. The temperature of the blackened end is 17°C when steady-state is reached. Stefen constant π = 6x10⁻⁸ W/m²-K⁴. Find the thermal conductivity of the material of the rod.
The figure for Q-48
Answer: Area of the blackened end, A =1 cm² =0.0001 m². The temperature of this end, T =17°C =17+273 =290 K. The temperature of the evacuated chamber, T' =27°C =27+273 =300 K. Hence the radiation being received by the black end,
Q/t =πA(T'⁴-T⁴)
=6x10⁻⁸*0.0001(300⁴-290⁴)
=0.0062 J/s
Now, this is the heat current in the steady-state in the rod. Let the conductivity of the rod = K. The temperature difference between the ends, ΞT =17°C =17 K. Length of the rod, L =50 cm =0.50 m. Cross-sectional area, A =0.0001 m².So the heat current in the rod is given as
Q/t =KAΞT/L
→0.0062 =K*0.0001*17/0.50
→K =62*0.50/17
=1.82 W/m-°C
49. One end of a rod of length 20 cm is inserted in a furnace at 800 K. The sides of the rod are covered with an insulating material and the other end emits radiation like a blackbody. The temperature of this end is 750 K in the steady-state. The temperature of the surrounding air is 300 K. Assuming radiation to be the only important mode of energy transfer between the surrounding and the open end of the rod, find the thermal conductivity of the rod. Stefen constant π = 6.0x10⁻⁸ W/m²-K⁴.
Answer: In the steady-state, the heat current in the rod is the same as radiation being lost to the surrounding through the blackbody end. Hence,
KA*ΞT/L =πA(T'⁴-T⁴)
→K =π(T'⁴-T⁴)L/ΞT
=6x10⁻⁸(750⁴-300⁴)*0.20/(800-750)
=74 W/m-K.
50. A calorimeter of negligible heat capacity contains 100 cc of water at 40°C. The water cools to 35°C in 5 minutes. The water is now replaced by K-oil of equal volume at 40°C. Find the time taken for the temperature to become 35°C under similar conditions. Specific heat capacities of water and K-oil are 4200 J/kg-K and 2100 J/kg-K respectively. Density of K-oil = 800 kg/m³.
Answer: Let the temperature of surrounding = T. Average temperature of water, T' = (40+35)/2 = 37.5 °C.
Average temperature difference from the surrounding =37.5-T
Heat lost by the water,
=0.10*4200(40-35)
=2100 J.
The average rate of heat loss =2100/5 J/min
=420 J/min.
If the conductivity of calorimeter = K, thickness =L and area =A, then
KA(37.5-T)/L =420 ---- (i)
In the case of K-oil,
mass, m =density*volume
= 0.08 kg,
Specific heat capacity, s= 2100 J/kg-K.
Heat loss, = 0.08*2100*(40-35)
=168*5 J
=840 J
If the K-oil takes t mins to cool then heat loss rate =840/t, Now
840/t = KA(37.5-T)/L -------- (ii)
From (i) and (ii)
840/t =420
→t =840/420 =2 min.
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CHAPTER- 26-Laws of Thermodynamics
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CHAPTER- 24-Kinetic Theory of Gases
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CHAPTER- 21 - Speed of Light
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CHAPTER- 16 - Sound Waves
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CHAPTER- 14 - Fluid Mechanics
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CHAPTER- 24-Kinetic Theory of Gases
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CHAPTER- 20 - Dispersion and Spectra
CHAPTER- 19 - Optical Instruments
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CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
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CHAPTER- 4 - The Forces
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CHAPTER- 2 - "Physics and Mathematics"
CHAPTER- 2 - "Physics and Mathematics"
