PHYSICS 231 Lecture 28: Thermal conduction ISOLATION Remco Zegers Walk-in hour: Thursday 11:30-13:30 am Helproom PHY 231

Previously: Phase Change GAS(high T) Q=cgasmT Q=csolidmT Solid (low T) liquid  solid Gas  liquid Q=mLv liquid (medium T) Q=mLf Q=cliquidmT PHY 231

PHY 231

How can heat be transferred? PHY 231

Conduction Touching different materials: Some feel cold, others feel warm, but all are at the same temperature… PHY 231

Thermal conductivity metal T=200C wood T=200C The heat transfer in the metal is much faster than in the wood: (thermal conductivity) T=370C T=370C PHY 231

Heat transfer via conduction Conduction occurs if there is a temperature difference between two parts of a conducting medium Rate of energy transfer P P=Q/t (unit Watt) P=kA(Th-Tc)/x=kAT/x k: thermal conductivity Unit:J/(msoC) metal k~300 J/(msoC) gases k~0.1 J/(msoC) nonmetals~1 J/(msoC) PHY 231

Example A glass window (A=4m2,x=0.5cm) separates a living room (T=200C) from the outside (T=0oC). A) What is the rate of heat transfer through the window (kglass=0.84 J/(msoC))? B) By what fraction does it change if the surface becomes 2x smaller and the temperature drops to -200C? A) P=kAT/x=0.84*4*20/0.005=13440 Watt B) Porig=kAT/x Pnew=k(0.5A)(2T)/x=Porig The heat transfer is the same PHY 231

Another one. Heat sink Heat reservoir An insulated gold wire (I.e. no heat lost to the air) is at one end connected to a heat reservoir (T=1000C) and at the other end connected to a heat sink (T=200C). If its length is 1m and P=200W what is its cross section (A)? kgold=314 J/(ms0C). P=kAT/x=314*A*80/1=25120*A=200 A=8.0E-03 m2 PHY 231

And another Water 0.5L 1000C A=0.03m2 thickness: 0.5cm. 1500C A student working for his exam feels hungry and starts boiling water (0.5L) for some noodles. He leaves the kitchen when the water just boils.The stove’s temperature is 1500C. The pan’s bottom has dimensions given above. Working hard on the exam, he only comes back after half an hour. Is there still water in the pan? (Lv=540 cal/g, kpan=1 cal/(ms0C) To boil away 0.5L (=500g) of water: Q=Lv*500=270000 cal Heat added by the stove: P=kAT/x=1*0.03*50/0.005= =300 cal P=Q/t t=Q/P=270000/300=900 s (15 minutes) He’ll be hungry for a bit longer… PHY 231

Isolation Tc Th inside L1 L2 L3 A house is built with 10cm thick wooden walls and roofs. The owner decides to install insulation. After installation the walls and roof are 4cm wood+2cm isolation+4cm wood. If kwood=0.10 J/(ms0C) and kisolation=0.02 J/(ms0C), by what factor does he reduce his heating bill? Pbefore=AT/[0.10/0.10]=AT Pafter=AT/[0.04/0.10+0.02/0.02+0.04/0.10]=0.55AT Almost a factor of 2 (1.81)! PHY 231

Overview of material for exam: 8 =I (compare to F=ma) Moment of inertia I: I=(miri2) (kgm2) : angular acceleration (rad/s2) I depends on the choice of rotation axis!! F r PHY 231

Chapter 8: Rotational KE Rotational KEr=½I2 Conservation of energy for rotating object: [PE+KEt+KEr]initial= [PE+KEt+KEr]final [mgh+½mv2+½I2]initial= [mgh+½mv2+½I2]final PHY 231

Ch 8: Angular momentum =v/r L=I =mr2v/r =mvr Conservation of angular momentum If the net torque equal zero, the angular momentum L does not change Iii=Iff PHY 231

Chapter 9 Young’s modulus. PHY 231

Chapter 9 Shear modulus PHY 231

Chapter 9: Bulk modulus PHY 231

Chapter 9: density water=1.0x103 kg/m3 PHY 231

Ch. 9 Pascal Enclosed fluid: F1/A1=F2/A2 Pascal’s principle: a change in pressure applied to a fluid that is enclosed is transmitted to the whole fluid and all the walls of the container that hold the fluid. PHY 231

h: distance between liquid surface and the point where you measure P h Pressure at depth h P = P0+ fluidgh h: distance between liquid surface and the point where you measure P h P Buoyant force for submerged object B = fluidVobjectg = Mfluidg = wfluid The buoyant force equals the weight of the amount of water that can be put in the volume taken by the object. If object is not moving: B=wobject object= fluid h B w Buoyant force for floating object The buoyant force equals the weight of the amount of water that can be put in the part of the volume of the object that is under water. objectVobject= waterVdisplaced h= objectVobject/(waterA) PHY 231

Bernoulli’s equation P1+½v12+gy1= P2+½v22+gy2 P+½v2+gy=constant The sum of the pressure (P), the kinetic energy per unit volume (½v2) and the potential energy per unit volume (gy) is constant at all points along a path of flow. Note that for an incompressible fluid: A1v1=A2v2 This is called the equation of continuity. PHY 231

Fs Fs Fg Surface tension Fs=L L: contact length between object and liquid : surface tension Fs Fs  Fg Units of : N/m=J/m2 Energy per unit surface PHY 231

Poiseuille’s Law How fast does a fluid flow through a tube? R4(P1-P2) (unit: m3/s) Rate of flow Q= v/t= 8L : coefficient of viscosity PHY 231

Ch. 10: Temperature scales Conversions Tcelsius=Tkelvin-273.5 Tfahrenheit=9/5*Tcelcius+32 We will use Tkelvin. If Tkelvin=0, the atoms/molecules have no kinetic energy and every substance is a solid; it is called the Absolute zero-point. Celsius Fahrenheit Kelvin PHY 231

Ch. 10: Thermal expansion L=LoT A=AoT =2 V=VoT =3  length L L=LoT surface A=AoT =2 volume V=VoT =3  L0 : coefficient of linear expansion different for each material Some examples: =24E-06 1/K Aluminum =1.2E-04 1/K Alcohol T=T0 T=T0+T PHY 231

Boyle & Charles & Gay-Lussac IDEAL GAS LAW PV/T = nR = Nkb n: number of particles in the gas (mol) R: universal gas constant 8.31 J/mol·K N: number of atoms/molecules kb:boltzmann’s constant 1.38x10-23 J/K n=N/NA NA:Avogadro’s constant 6.02x1023 If no molecules are extracted from or added to a system: PHY 231

Pressure Number of Molecules Mass of 1 molecule Averaged squared velocity Number of molecules per unit volume Volume Average translation kinetic energy PHY 231

Average molecular kinetic energy Ideal gases Average molecular kinetic energy Total kinetic energy rms speed of a molecule M=Molar mass (kg/mol) PHY 231

Ch. 11 Heat transfer to an object The amount of energy transfer Q to an object with mass m when its temperature is raised by T: Q=cmT Change in temperature Mass of object Energy transfer (J or cal) Specific heat (J/(kgoC) or cal/(goC) PHY 231

Calorimetry If we connect two objects with different temperature energy will transferred from the hotter to the cooler one until their temperatures are the same. If the system is isolated: Qcold=-Qhot mcoldccold(Tfinal-Tcold)=-mhotchot(Tfinal-Thot) the final temperature is: Tfinal= mcoldccoldTcold+mhotchotThot mcoldccold+mhotchot PHY 231

Phase Change GAS(high T) Q=cgasmT Q=csolidmT Solid (low T) liquid  Q=mLv liquid (medium T) Q=mLf Q=cliquidmT Make sure to understand: ice -> water -> steam PHY 231

Heat transfer via conduction Rate of energy transfer P P=Q/t (unit Watt) P=kA(Th-Tc)/x=kAT/x k: thermal conductivity Unit:J/(msoC) For several layers: PHY 231