Physics 31 Lecture 31 Mi Main pints f tday s lecture: Heat and heat capacity: Q = cmδt Phase transitins and latent heat: Q = LΔm Mechanisms f heat flw. Cnductive heat flw ΔQ kat ( T1 ) H = = Δt L Examples f heat cnductivity, R values fr insulatrs R = L / k i i i
Heat and heat capacity We saw that fr atms r mlecules in an ideal gas <KE>=3/k B T. In general, the atms r mlecules in matter increase in energy if the bject is heated. This is thermal energy, which can be transfer t and frm the gas. We define the heat energy Q t be the energy that flws frm a ht bject t a cld bject slely because f the difference in T. When the heat flw is sufficient, i the tw bjects will reach the same temperature and we say they are in thermal equilibrium at the same T. If we put Q int a mn-atmic ideal gas at cnstant vlume: 3 3 3 Q= Δ Eth = Eth,f Eth,i = Nkb Tf Nkb Ti = Nkb( Tf Ti) 3 3 Q= Nk c mlar,v =3/R is the heat capacity per mle at bδt = nr Δ T = ncmlar,vδt cnstant vlume fr mn-atmic ideal gas Fr many ther materials, the relatinship between heat transferred and temperature change is given by: (usually transferred at cnstant V r P) Q = cmδt c is the specific heat capacity per unit mass m is the mass and cm the ttal heat capacity f the bject.
Example Hli Helium (He), (H) a mnatmicgas, fills a 0.010 010 m 3 cntainer. ti The pressure f the gas is 6.x10 5 Pa. Hw lng wuld a 0.5 hp engine have t run (1 hp=746 W) t prduce an amunt f wrk equal t the th thermal energy f this gas? 3 Eth = N KE = NkT = B PV = nrt 3 nrt 3 3 = 6.x10 Pa 0.010m = 9300J Eth = PpressureV ( 5 )( 3) ( ) E = Wrk = 0.5 746W Δt E 9300J Δ t = = 50s 0.5 746W 187W ( )
Quiz Tw mles f fthe mnatmic gas helium (C V =3/R) are initially iti at a temperature f 300K. The gas is cled at cnstant vlume. Hw much heat must be remved t decrease the temperature (in Kelvin) by a factr f tw? (Nte: R=N A k B =8.31 J/(mle k)) a) 1.7 kj 3 3 b).7 kj Eth,i = nrt E i th,f = nrtf c) 3.7 kj Eth,f = Eth,i + Q d) 4.7 kj 3 3 e) 5.7 kj Q= Eth,f E 1 th,i = nrtf nrt i TF = Ti 3 Q nr( Tf Ti) = 3 mles 8.31J / mle K 150K = ( ) ( ) ( ) Q = 3.7kJ remved energy = 3.7 kj
Cnservatin f heat energy Imagine tw bjects A and B, with thermal energies E th,a.0 and E th,b,0 and T A >T B. When they are put in thermal cntact, they will cme t equilibrium and heat Q will be transferred frm A t B. Then, we can write Eth,A,f = Eth,A,0 + QA Eth,B,f = Eth,B,0 + QB Frm cnservatin f energy: Q + Q = 0 Q = Q = Q B A B A E = E Q Eth,B,f = Eth,B,0 + Q th,a,f th,a,0
Example f equilibratin and energy cnservatin At a fabricatin plant, a ht frging has a mass f 75 kg and a specific heat capacity f 430 J/(kg C). T harden it, the frging is quenched by immersin in 710 kg f il that has a temperature f 3 C and a specific heat capacity f 645 cal/(kg C). The final temperature f the il and frging at thermal equilibrium i is 47 C. Assuming that theat flws nly between the frging and the il, determine the initial temperature f the frging. Q = c m T T Qil = cilmil ( Tf Til ) ( ) f Energy cnservatin: Q + Q = 0 il Q = Q 419J 4.19J cil = 645cal / (kg C) = 700kJ / (kg C) cal ( ) = ( ) c m T T c m T T T f il il f il f T = T = T + f ( ) c m T T il il f il cm c m T T ( ) il il f il c m il m m il c c il T f 75 kg 710 kg 430J/kg/ C T il,0 3 C T,0? ( )( ) C) ( 75kg) 645cal/kg/ C 47 C = + = 430J / (kg 700J / (kg C) 710kg 47 3 C 47 C 939 C
Example Example: A 100 kg mass f water at a temperature f 30 C is drpped int a thermally islated vessel cntaining 50 kg f water which is a temperature f 10 C. After system cmes t thermal equilibrium the final temperature is (useful infrmatin: i c water =4186 J/(kg C)) a) 11 C Q ( ) b) 14 C 100 = cwater100kg Tf 30 C c) 19 C Q ( ) 50 = cwater50kg Tf 10 C d) 3 C Q100 + Q50 = 0 ( ) + ( ) c 100kg T 30 C + c 50kg T 10 C = 0 water f water f ( ) f T 30 C + Tf 10 C = 0 Tf = 3 C 3Tf = 60 C + 10 C
Phase transitins T Many substances display a plateaus in the calric curve that describes T vs. Q. Regins with a linear increase crrespnd t a cnstant heat capacity where ΔQ = cmδt ΔT Q = ΔQ /(cm) The flat regins ccur when the system has a phase transitin. Fr water, there is ne where ice changes t water and ther where water changes t steam. If the pressure remains cnstant during the phase change, the temperatures will remain cnstant. The heat required t change a mass Δm f the matter is given by the latent heat L fr the phase change. ΔQ = LΔm 4 5 Lfusin = 33.5x10 J / kg Lvaprizati n =.6x10 J / kg
Example When it rains, water vapr in the air cndenses int liquid id water, and the energy is released. (a) Hw much energy is released when 0.054 m (ne inch) f rain falls ver an area f.59x10 6 m (ne square mile)? (b) If the average energy needed t heat ne hme fr a year is 1.5x10 11 J, hw many hmes can be heated with the energy determined in part (a)? a) m ρ V = ( ) ( ) water water water = 1000kg / m 3.59x10 6 m.054m = 6.58x10 7 kg Δ Qreleased = Lvaprizatinm ( )( ) water =.6x10 5 J / kg 6.58x10 7 kg = 1.5x10 14 J huses 11 14 n = 1000 b) n 1.5x10 J = 1.5x10 J huses
Heat flw There are three majr prcesses that t transfer heat frm ne pint t anther. Cnductin Cnvectin Radiatin Cnvectin results frm the fact that ht bjects generally expand. This decreases their density. If this ccurs in a fluid, the less dense ht fluid rises and the clder denser fluid falls.
Cnductin Cnductin cncerns the transfer f heat thrugh h materials withut t cnvectin. Cnsider an cncrete wall f a heated garage. The utside f the garage is at temperature T 1 and the interir f the garage is a temperature T. The cnductive heat flw thrugh a prtin f the wall with area A is given by: ΔQ ka( T T1 ) kaδt H = = = Δt L L H is the heat flw thrugh the wall, k is the thermal cnductivity f cncrete, and L is the thickness f the cncrete wall. Example: Calculate the heat flw thrugh a m sectin f a 0 cm thick cncrete wall when the utside temperature is 0 C and the inside temperature is 0 C. Assume the thermal cnductive f the cncrete is 1.3 J/(s m C) Δ Q H = = Δt ka Δ T L ( ) ( ) 13J/(s 1.3J m C) m 0 C = = 60W 0.m