Raymond A. Serway Chris Vuille Chapter Eleven Energy in Thermal Processes Energy Transfer •  When two objects of different temperatures are placed in thermal contact, the temperature of the warmer decreases and the temperature of the cooler increases •  The energy exchange ceases when the objects reach thermal equilibrium •  The concept of energy was broadened from just mechanical to include internal –  Made ConservaGon of Energy a universal law of nature IntroducGon Heat Compared to Internal Energy •  Internal Energy, U, is the energy associated with the atoms and molecules of the system •  Heat is the transfer of energy, Q , between a system and its environment because of a temperature difference between them. SecGon 11.1 Units of Heat •  Calorie –  A calorie is the amount of energy necessary to raise the temperature of 1 g of water from 14.5° C to 15.5° C . •  A Calorie (food calorie) is 1000 cal •  1 cal = 4.186 J –  This is called the Mechanical Equivalent of Heat SecGon 11.1 Working Off Breakfast: A student eats a breakfast consisting of a bowl of
cereal and milk, containing a total of 3.20 × 102 Calories of energy. He
wishes to do an equivalent amount of work in the gymnasium by performing
curls with a 25.0-kg barbell. How many times must he raise the weight to
expend that much energy? Assume he raises it through a vertical
displacement of 0.400 m each time, the distance from his lap to his upper
chest.
Working Off Breakfast: A student eats a breakfast consisting of a bowl of
cereal and milk, containing a total of 3.20 × 102 Calories of energy. He
wishes to do an equivalent amount of work in the gymnasium by performing
curls with a 25.0-kg barbell. How many times must he raise the weight to
expend that much energy? Assume he raises it through a vertical
displacement of 0.400 m each time, the distance from his lap to his upper
chest.
Q = 3.2 × 10 2 Cal = 3.2 × 10 2 kcal = 3.2 × 10 2 ( 4,186 J) = 1339520. J
Wtotal = N2mgh = Q ⇒ N =
€
Q
1339520. J
=
= 6834
2mgh 2(25.0kg)(9.8m/s2 )(0.4m)
James PrescoW Joule •
•
•
•
1818 – 1889 BriGsh physicist ConservaGon of Energy RelaGonship between heat and other forms of energy transfer SecGon 11.1 Heat Capacity •  The heat capacity of an object is the amount of heat added to it divided by its rise in temperature: Q is posiGve if ΔT is posiGve; that is, if heat is added to a system. Q is negaGve if ΔT is negaGve; that is, if heat is removed from a system. SecGon 11.2 Specific Heat •  The heat capacity of an object depends on its mass. A quanGty which is a property only of the material is the specific heat: SecGon 11.2 Units and Values of Specific Heat •  SI units –  J / kg °C •  Historical units –  cal / g °C •  See table 11.1 for the specific heats of various materials –  These values are typical •  They may vary depending on the temperature and state of the material SecGon 11.2 Specific Heats Here are some specific heats of various materials: Heat and Specific Heat •  Q = m c ΔT •  ΔT is always the final temperature minus the iniGal temperature •  When the temperature increases, ΔT and ΔQ are posiGve and energy flows into the system •  When the temperature decreases, ΔT and ΔQ are negaGve and energy flows out of the system SecGon 11.2 A Consequence of Different Specific Heats •  Water has a high specific heat compared to land •  On a hot day, the air above the land warms faster •  The warmer air flows upward and cooler air moves toward the beach SecGon 11.2 Calorimeter •  Used in one technique for determining the specific heat of a substance •  A calorimeter is a vessel that is a good insulator which allows a thermal equilibrium to be achieved between substances without any energy loss to the environment SecGon 11.3 Calorimetry •  Analysis performed using a calorimeter •  ConservaGon of energy applies to the isolated system •  The energy that leaves the warmer substance equals the energy that enters the water –  Qcold = -­‐Qhot –  NegaGve sign keeps consistency in the sign convenGon of ΔT SecGon 11.3 Calorimetry with More Than Two Materials •  In some cases it may be difficult to determine which materials gain heat and which materials lose heat •  You can start with ΣQk = 0 –  Qk is the energy of the kth object –  Each Q = m c ΔT –  Use Tf – Ti –  You don’t have to determine before using the equaGon which materials will gain or lose heat SecGon 11.3 Stressing a Strut: A steel strut near a ship’s furnace is 2.00 m long, with a
mass of 1.57 kg and cross-sectional area of 1.00 × 1024 m2. During operation
of the furnace, the strut absorbs a net thermal energy of 2.50 × 105 J. (a)
Find the change in temperature of the strut. (b) Find the increase in length of
the strut. (c) If the strut is not allowed to expand because it’s bolted at each
end, find the compressional stress developed in the strut.
(a)
Q
2.5 × 10 5 J

Q = cmΔT ⇒ ΔT =
=
=
355
C

cm ( 448 J/kg C)(1.57kg)
(b) ΔL = αL0 ΔT = (11 × 10 −6 C)(2.00 m)( 355  C) = 7.8 × 10 −3 m
F
7.8 × 10 −3 m
ΔL
−11
(c)
=Y
= (2. × 10 Pa)
= 7.8 × 10 8 Pa
A
2.01m
L0
€
Finding a Specific Heat: A 125-g block of an unknown substance with a
temperature of 90.0°C is placed in a Styrofoam cup containing 0.326 kg of
water at 20.0°C. The system reaches an equilibrium temperature of 22.4°C.
What is the specific heat, cx, of the unknown substance if the heat capacity of
the cup is neglected?
Energy Q transferred from the substance to the water is:
Qx → H 2 0 = c x mx (22.4  C − 90.0  C)
Energy Q absorbed by the water is :
Q x → H 2 0→ x = c H 2 0 m H 2 0 (22.4  C − 20.0  C)
∑Q
cx =
€
k
= 0 ⇒ Qx → H 2 0 + Q H 2 0→ x = 0 ⇒ c x mx (22.4  C − 90.0  C) + c H 2 0 m H 2 0 (22.4  C − 20.0  C) = 0 ⇒
c H 2 0 m H 2 0 (22.4  C - 20. C)
mx (90. C - 22.4  C)
0.326kg(22.4  C - 20. C)
= ( 4190J /kgC )
= 388 J/kg C


0.125kg(90. C - 22.4 C)
Calculate an Equilibrium Temperature: Suppose 0.400 kg of water initially
at 40.0°C is poured into a 0.300-kg glass beaker having a temperature of
25.0°C. A 0.500-kg block of aluminum at 37.0°C is placed in the water and
the system insulated. Calculate the final equilibrium temperature of the
system.
Water - glass beaker - aluminum
∑Q
k
= 0 →QH 2 0 + Qglass + QAl = 0 ⇒
(
)
(
)
(
)
c H 2 0 m H 2 0 Tf − 40.0  C + c glassmglass Tf − 25  C + c Al mAl Tf − 37.0  C = 0 ⇒
(4186J/kg C)(0.4kg)(T

f
)
(
Tf (1674.40 + 251.1 + 450.) J/C - (66976. + 6277.5 +16650.) J = 0 ⇒
Tf = 37.9  C
€
)
(
)
− 40.0  C + (837J/kg C)(0.3kg) Tf − 25  C + (900J/kg C)(0.5kg) Tf − 37  C = 0 ⇒
Phase Changes •  A phase change occurs when the physical characterisGcs of the substance change from one form to another •  Common phases changes are –  Solid to liquid – melGng (fusion) –  Liquid to Solid – freezing (fusion) –  Gas to Liquid – condensaGon (vaporizaGon) –  Liquid to gas – boiling (vaporizaGon) •  Phases changes involve a change in the internal energy, but no change in temperature SecGon 11.4 Latent Heat •  The energy Q needed to change the phase of a given pure substance is –  Q = ±m L •  L is the called the latent heat of the substance –  Latent means hidden –  L depends on the substance and the nature of the phase change •  Choose a posiGve sign if you are adding energy to the system and a negaGve sign if energy is being removed from the system SecGon 11.4 Latent Heat, cont. •  SI unit of latent heat are J / kg •  Latent heat of fusion, Lf, is used for melGng or freezing •  Latent heat of vaporiza>on, Lv, is used for boiling or condensing •  Table 11.2 gives the latent heats for various substances SecGon 11.4 Latent Heat, cont. SecGon 11.4 SublimaGon •  Some substances will go directly from solid to gaseous phase –  Without passing through the liquid phase •  This process is called sublima>on –  There will be a latent heat of sublimaGon associated with this phase change SecGon 11.4 Graph of Ice to Steam SecGon 11.4 Warming Ice •  Start with one gram of ice at –30.0° C •  During A, the temperature of the ice changes from –30.0° C to 0º C •  Use Q = m c ΔT •  Will add 62.7 J of energy SecGon 11.4 MelGng Ice •  Once at 0° C, the phase change (melGng) starts •  The temperature stays the same although energy is sGll being added •  Use Q = m Lf •  Needs 333 J of energy SecGon 11.4 Warming Water •  Between 0° C and 100° C, the material is liquid and no phase changes take place •  Energy added increases the temperature •  Use Q = m c ΔT •  419 J of energy are added SecGon 11.4 Boiling Water •  At 100° C, a phase change occurs (boiling) •  Temperature does not change •  Use Q = m Lv •  2 260 J of energy are needed SecGon 11.4 HeaGng Steam •  Aqer all the water is converted to steam, the steam will heat up •  No phase change occurs •  The added energy goes to increasing the temperature •  Use Q = m c ΔT •  To raise the temperature of the steam to 120°, 40.2 J of energy are needed SecGon 11.4 Ice Water: At a party, 6.00 kg of ice at -5.00°C is added to a cooler holding 30
liters of water at 20.0°C. What is the temperature of the water when it comes to
equilibrium?
L f ,ice = 3.33 × 10 5 J/kg
c ice = 2090J/kg C
c water = 4186J/kg C
€
Ice Water: At a party, 6.00 kg of ice at -5.00°C is added to a cooler holding 30
liters of water at 20.0°C. What is the temperature of the water when it comes to
equilibrium?
L f ,ice = 3.33 × 10 5 J/kg
c ice = 2090J/kg C
c water = 4186J/kg C
m
ρ water = water ⇒ mwater = ( 30 × 10 −3 m3 )(1000kg/m3 ) = 30kg
Vwater
ice − water
∑Q
k
= 0 ⇒ Qice + Qmelt + Qice,water + Qwater = 0 ⇒
c ice mice (Tf + 5  C) + L f ,ice mice + c water mice (Tf − 0  C) + c water mwater (Tf − 20  C) = 0 ⇒



(2090J/kg C)(6.00kg)(0 C + 5 C) + (3.33 × 10 J /kg)(6.00kg) +
(4186J/kg C)(6kg)(T − 0 C) + (4186J/kg C)(30kg)(T − 20 C) = 0 ⇒


5

f
62700J + 2. × 10 6 J - 2.511 × 10 6 J + Tf (150696J/C) ⇒
Tf = 3 C
€

f
Partial Melting: A 5.00-kg block of ice at 0°C is added to an insulated
container partially filled with 10.0 kg of water at 15.0°C. (a) Find the final
temperature, neglecting the heat capacity of the container. (b) Find the mass
of the ice that was melted.
L f ,ice = 3.33 × 10 5 J/kg
c ice = 2090J/kg C
c water = 4186J/kg C
€
Partial Melting: A 5.00-kg block of ice at 0°C is added to an insulated
container partially filled with 10.0 kg of water at 15.0°C. (a) Find the final
temperature, neglecting the heat capacity of the container. (b) Find the mass
of the ice that was melted.
L f ,ice = 3.33 × 10 5 J/kg
c ice = 2090J/kg C
c water = 4186J/kg C
The total amount of energy available to be transferred to the ice to melt it is:
Q available = c water mwater (15  C) = ( 4186J/kg C)(10kg)(15  C) = 6.279 × 10 5 J
The amount of energy required to melt 5 kg of ice is :
Q melt = L f ,ice (5kg) = ( 3.33 × 10 5 J/kg)(5kg) = 1.665 × 10 6 J
There is not enough energy available so the ice will only partly melt and the
final temperature will be T = 0  C since ice melts at this temperature.
mmelt
€
Q available
6.279 × 10 5 J
=
=
= 1.89kg (this is the amount of ice that will be melted.)
3.33 × 10 5 J/kg
L f ,ice
Armageddon!: A comet half a kilometer in radius consisGng of ice at 273 K hits Earth at a speed of 4.00 × 104 m/s. For simplicity, assume all the kineGc energy converts to thermal energy on impact and that all the thermal energy goes into warming the comet. (a) Calculate the volume and mass of the ice. (b) Use conservaGon of energy to find the final temperature of the comet material. Assume, contrary to fact, that the result is superheated steam and that the usual specific heats are valid, although in fact they depend on both temperature and pressure. (c) Assuming the steam retains a spherical shape and has the same iniGal volume as the comet, calculate the pressure of the steam using the ideal gas law. This law actually doesn’t apply to a system at such high pressure and temperature, but can be used to get an esGmate. L f = 3.33 × 10 5 J/kg
L v = 2.26 × 10 6 J/kg
c ice = 2090J/kg C
c water = 4186J/kg C
c steam = 2010J/kg C
(a) Vcomet =
ρice =
€
3
4 3 4
πR = π (0.5 × 10 3 m) = 5.24 × 10 8 m3
3
3
mcomet
⇒ mcoment = ρiceVcomet = 4.8 × 1011 kg
Vcomet
Armageddon!: A comet half a kilometer in radius consisGng of ice at 273 K hits Earth at a speed of 4.00 × 104 m/s. For simplicity, assume all the kineGc energy converts to thermal energy on impact and that all the thermal energy goes into warming the comet. (a) Calculate the volume and mass of the ice. (b) Use conservaGon of energy to find the final temperature of the comet material. Assume, contrary to fact, that the result is superheated steam and that the usual specific heats are valid, although in fact they depend on both temperature and pressure. (c) Assuming the steam retains a spherical shape and has the same iniGal volume as the comet, calculate the pressure of the steam using the ideal gas law. This law actually doesn’t apply to a system at such high pressure and temperature, but can be used to get an esGmate. ⎛ 1 ⎞
2
1
2
= ⎜ ⎟( 4.8 × 1011 kg)( 4 × 10 4 m /s) = 3.84 × 10 20 J
(b) The energy available to heat the comet is : KE = micev comet
⎝ 2 ⎠
2
∑Q
k
= KE ⇒ Q ice + Qmelt + Qwater + Qvap + Qsteam = KE ⇒
(
)
c ice mcomet (0  C − (−0.15  C)) + mcomet L f + c water mcomet (100  C − 0  C) + mcomet Lv + c steam mcomet Tf −100  C = KE ⇒

11

11

11

(2090J/kg C)(4.8 × 10 kg)(0.15 C) + (4.8 × 10 kg)(3.33 × 10 J/kg) +
(4186J/kg C)(4.8 × 10 kg)(100 C) + (4.8 × 10 kg)(2.26 × 10 J/kg) +
(2010J/kg C)(4.8 × 10 kg)(T −100 C) ⇒
11

11
5
6

f
(
)
1.51 × 1014 J +1.6 × 1017 J + 2.0 × 1017 J +1.09 × 1018 J + 9.65 × 1014 Tf −100  C = 3.84 × 10 20 J ⇒
(
)
Tf −100  C =
€
3.83 × 10 20 J
5
5 
C
14 = 3.96 × 10 C ⇒ Tf ≅ 4 × 10
9.65 × 10
Armageddon!: A comet half a kilometer in radius consisGng of ice at 273 K hits Earth at a speed of 4.00 × 104 m/s. For simplicity, assume all the kineGc energy converts to thermal energy on impact and that all the thermal energy goes into warming the comet. (a) Calculate the volume and mass of the ice. (b) Use conservaGon of energy to find the final temperature of the comet material. Assume, contrary to fact, that the result is superheated steam and that the usual specific heats are valid, although in fact they depend on both temperature and pressure. (c) Assuming the steam retains a spherical shape and has the same iniGal volume as the comet, calculate the pressure of the steam using the ideal gas law. This law actually doesn’t apply to a system at such high pressure and temperature, but can be used to get an esGmate. 13
5
nRT (2.66 × 10 mol)(8.31J/molK)( 4 × 10 K)
11
=
=
1.69
×
10
Pa
(c) PV = nRT ⇒ P =
8
3
5.24 × 10 m
V
mcomet
4.8 × 1011 kg
where n =
=
= 2.66 × 1013 mol
−3
molar mass 18.02 × 10 kg/mol
€
Methods of Heat Transfer •  Methods of Heat Transfer include –  ConducGon –  ConvecGon –  RadiaGon SecGon 11.5 ConducGon •  The transfer can be viewed on an atomic scale –  It is an exchange of energy between microscopic parGcles by collisions –  Less energeGc parGcles gain energy during collisions with more energeGc parGcles •  Rate of conducGon depends upon the characterisGcs of the substance SecGon 11.5 ConducGon example •  The molecules vibrate about their equilibrium posiGons •  ParGcles near the stove coil vibrate with larger amplitudes •  These collide with adjacent molecules and transfer some energy •  Eventually, the energy travels enGrely through the pan and its handle SecGon 11.5 ConducGon, cont. •  The rate of conducGon depends on the properGes of the substance •  In general, metals are good conductors –  They contain large numbers of electrons that are relaGvely free to move through the metal –  They can transport energy from one region to another •  ConducGon can occur only if there is a difference in temperature between two parts of the conducGng medium SecGon 11.5 ConducGon, equaGon •  The slab of material allows energy Q to transfer from the region of higher temperature to the region of lower temperature •  A is the cross-­‐secGonal area SecGon 11.5 ConducGon, equaGon explanaGon •  A is the cross-­‐secGonal area •  Through a rod, Δx = L •  P is in WaWs when Q is in Joules and t is in seconds •  k is the thermal conduc>vity of the material –  See table 11.3 for some conducGviGes –  Good conductors have high k values and good insulators have low k values SecGon 11.5 ConducGon Is the heat conducted in the parallel arrangement (a) greater than, (b) less than, or (c) the same as the heat conducted with the rods in series. ConducGon SecGon 11.5 Home InsulaGon •  Substances are rated by their R values –  R = L / k –  See table 11.4 for some R values •  For mulGple layers, the total R value is the sum of the R values of each layer •  Wind increases the energy loss by conducGon in a home SecGon 11.5 ConducGon and InsulaGon with MulGple Materials •  Each porGon will have a specific thickness and a specific thermal conducGvity •  The rate of conducGon through each porGon is equal SecGon 11.5 MulGple Materials, cont. •  The rate through the mulGple materials will be •  TH and TC are the temperatures at the outer extremiGes of the compound material SecGon 11.5 Conduc.ve Losses from the Human Body: In a human being, a layer of fat and muscle lies under the skin having various thicknesses depending on locaGon. In response to a cold environment, capillaries near the surface of the body constrict, reducing blood flow and thereby reducing the conducGvity of the Gssues. These Gssues form a shell up to an inch thick having a thermal conducGvity of about 0.21 W/m K, the same as skin or fat. (a) EsGmate the rate of loss of thermal energy due to conducGon from the human core region to the skin surface, assuming a shell thickness of 2.0 cm and a skin temperature of 33.0°C. (b) Calculate the thermal energy lost due to conducGon in 1.0 h. (c) EsGmate the change in body temperature in 1.0 h if the energy is not replenished. Assume a body mass of 75 kg and a skin surface area of 1.73 m2. Conduc.ve Losses from the Human Body: In a human being, a layer of fat and muscle lies under the skin having various thicknesses depending on locaGon. In response to a cold environment, capillaries near the surface of the body constrict, reducing blood flow and thereby reducing the conducGvity of the Gssues. These Gssues form a shell up to an inch thick having a thermal conducGvity of about 0.21 W/m K, the same as skin or fat. (a) EsGmate the rate of loss of thermal energy due to conducGon from the human core region to the skin surface, assuming a shell thickness of 2.0 cm and a skin temperature of 33.0°C. (b) Calculate the thermal energy lost due to conducGon in 1.0 h. (c) EsGmate the change in body temperature in 1.0 h if the energy is not replenished. Assume a body mass of 75 kg and a skin surface area of 1.73 m2. (a) P =
(0.21 W /m K)(1.73m2 )(37  C − 33 C)
2 × 10 -2 m
= 73W
(b) Q = PΔt = ( 73W )( 3600s) = 2.6 × 10 5 J
(c) Q = c body mbody (Tf − 37  C) = 2.6 × 10 5 J ⇒
2.6 × 10 5 J
2.6 × 10 5 J
=
⇒ (Tf − 37  C) = 1.0  C
(Tf − 37 C) =
c body mbody ( 3470J/kg/K)( 75kg)

€
Conduction
Windows A 0.5 cm thick glass window in a house is 1.0 m on a side. (a) How much heat is lost through this window in 1 day if the temperature in the house is 21 °C and the temperature outside is 0.0 °C ? (b) Suppose the area is quadrupled and the thickness is doubled. If everything else is the same by what factor does the heat flow change? €
Conduction
Windows A 0.5 cm thick glass window in a house is 1.0 m on a side. (a) How much heat is lost through this window in 1 day if the temperature in the house is 21 °C and the temperature outside is 0.0 °C ? (b) Suppose the area is quadrupled and the thickness is doubled. If everything else is the same by what factor does the heat flow change? (a) The flow of heat out of the window is :
(0.84W/m/K)(1m2 )(21 C − 0 C)
ΔT
t=
(24 × 3600 s)
Q = kA
0.005m
L
Q = 3.0 × 10 8 J
(b) If you quadruple the area Q increases by 4 but if you double
the thickness Q is reduced by a factor of 2. So the Q increases by 2.
Construc.on and Thermal Insula.on: (a) Find the energy transferred in 1.00 h by conducGon through a concrete wall 2.0 m high, 3.65 m long, and 0.20 m thick if one side of the wall is held at 5.00°C and the other side is at 20.0°C (Fig. 11.8). Assume the concrete has a thermal conducGvity of 0.80 J/s m °C. (b) The owner of the home decides to increase the insulaGon, so he installs 0.50 in of thick sheathing, 3.5 in of fiberglass bavng, and a drywall 0.50 in thick. Calculate the R-­‐
factor. (c) Calculate the energy transferred in 1.00 h by conducGon. Construc.on and Thermal Insula.on: (a) Find the energy transferred in 1.00 h by conducGon through a concrete wall 2.0 m high, 3.65 m long, and 0.20 m thick if one side of the wall is held at 5.00°C and the other side is at 20.0°C (Fig. 11.8). Assume the concrete has a thermal conducGvity of 0.80 J/s m °C. (b) The owner of the home decides to increase the insulaGon, so he installs 0.50 in of thick sheathing, 3.5 in of fiberglass bavng, and a drywall 0.50 in thick. Calculate the R-­‐factor. (c) Calculate the energy transferred in 1.00 h by conducGon P=
Q A(Th − Tc ) A(Th − Tc )
=
=
t
R
∑ L i /k i
⎛
0.2m
m2 ⎞
m2
R=
+ 2⎜0.030
 ⎟ = 0.31

J/s/m/
C ⎠
J/s/m/C
0.8J/s/m/
C
⎝
(
)
(a) Q =
(2.0m × 3.65m)(20  C − 5  C)( 3600s)
2
0.31
€
m
J/s/m/C
= 1.3 × 10 6 J
Construc.on and Thermal Insula.on: (a) Find the energy transferred in 1.00 h by conducGon through a concrete wall 2.0 m high, 3.65 m long, and 0.20 m thick if one side of the wall is held at 5.00°C and the other side is at 20.0°C (Fig. 11.8). Assume the concrete has a thermal conducGvity of 0.80 J/s m °C. (b) The owner of the home decides to increase the insulaGon, so he installs 0.50 in of thick sheathing, 3.5 in of fiberglass bavng, and a drywall 0.50 in thick. Calculate the R-­‐
factor. (c) Calculate the energy transferred in 1.00 h by conducGon. (b) R total = R air -layer + R concrete + R sheath + R fiberglass + R drywall + R air -layer ⇒
R total
m2
= (0.030 + 0.25 + 0.233 +1.92 + 0.079 + 0.030) = 2.5
J/s/m/C
(c) Q =
(2.0m × 3.65m)(20  C − 5  C)( 3600s)
2
2.5
€
m
J/s/m/C
= 1.6 × 10 5 J
ConvecGon •  Energy transferred by the movement of a substance –  When the movement results from differences in density, it is called natural convec>on –  When the movement is forced by a fan or a pump, it is called forced convec>on SecGon 11.5 ConvecGon example •  Air directly above the flame is warmed and expands •  The density of the air decreases, and it rises •  The mass of air warms the hand as it moves by SecGon 11.5 ConvecGon applicaGons •
•
•
•
•
Boiling water Radiators Upwelling Cooling automobile engines Algal blooms in ponds and lakes SecGon 11.5 ConvecGon Current Example •  The radiator warms the air in the lower region of the room •  The warm air is less dense, so it rises to the ceiling •  The denser, cooler air sinks •  A conGnuous air current paWern is set up as shown SecGon 11.5 RadiaGon •  RadiaGon does not require physical contact •  All objects radiate energy conGnuously in the form of electromagneGc waves due to thermal vibraGons of the molecules •  Rate of radiaGon is given by Stefan’s Law SecGon 11.5 RadiaGon example •  The electromagneGc waves carry the energy from the fire to the hands •  No physical contact is necessary •  Cannot be accounted for by conducGon or convecGon SecGon 11.5 RadiaGon equaGon •  P = σ A e T4 –  The power is the rate of energy transfer, in WaWs –  σ = 5.669 6 x 10-­‐8 W/m2.K4 •  Called the Stefan-­‐Boltzmann constant –  A is the surface area of the object –  e is a constant called the emissivity •  e varies from 0 to 1 –  T is the temperature in Kelvins SecGon 11.5 Energy AbsorpGon and Emission by RadiaGon •  The rate at which the object at temperature T with surroundings at To radiates is –  Pnet = σ A e (T4 -­‐ To4) –  When an object is in equilibrium with its surroundings, it radiates and absorbs at the same rate •  Its temperature will not change SecGon 11.5 Ideal Absorbers •  An ideal absorber is defined as an object that absorbs all of the energy incident on it –  e = 1 •  This type of object is called a black body •  An ideal absorber is also an ideal radiator of energy SecGon 11.5 Ideal Reflector •  An ideal reflector absorbs none of the energy incident on it –  e = 0 SecGon 11.5 ApplicaGons of RadiaGon •  Clothing –  Black fabric acts as a good absorber –  White fabric is a beWer reflector •  Thermography –  The image of the paWern formed by varying radiaGon levels is called a thermogram •  Body temperature –  RadiaGon thermometer measures the intensity of the infrared radiaGon from the eardrum SecGon 11.5 ResisGng Energy Transfer •  Dewar flask/thermos boWle •  Designed to minimize energy transfer to surroundings •  Space between walls is evacuated to minimize conducGon and convecGon •  Silvered surface minimizes radiaGon •  Neck size is reduced SecGon 11.5 Global Warming •  Greenhouse example –  Visible light is absorbed and re-­‐emiWed as infrared radiaGon •  Earth’s atmosphere is also a good transmiWer of visible light and a good absorber of infrared radiaGon SecGon 11.6 Polar Bear Club: A member of the Polar Bear Club, dressed only in bathing
trunks of negligible size, prepares to plunge into the Baltic Sea from the beach
in St. Petersburg, Russia. The air is calm, with a temperature of 5°C. If the
swimmer’s surface body temperature is 25°C, compute the net rate of energy
loss from his skin due to radiation. How much energy is lost in 10.0 min?
Assume his emissivity is 0.900 and his surface area is 1.50 m2.
Polar Bear Club: A member of the Polar Bear Club, dressed only in bathing
trunks of negligible size, prepares to plunge into the Baltic Sea from the beach
in St. Petersburg, Russia. The air is calm, with a temperature of 5°C. If the
swimmer’s surface body temperature is 25°C, compute the net rate of energy
loss from his skin due to radiation. How much energy is lost in 10.0 min?
Assume his emissivity is 0.900 and his surface area is 1.50 m2.
[
4
4
]
(a) P = σAe(T 4 − T04 ) = (5.67 × 10 −8 W /m 2K 4 )(1.5m 2 )(0.90) (298K) − (278K) ⇒
P = 146W
(b) Q = PΔt = (146W )(600s) = 8.76 × 10 4 J
€