Specific Heat, Latent Heat - Physics - The Citadel

Physics 204 – College Physics II Department of Physics – The CitadelFriday’s ReadingIf you haven’t finished chapter 14, please do so. It is allimportant.Today we will discuss sections mostly sections 3 – 5 onspecific and latent heats.The key concepts include calories, specific heat, and latentheat.I plan to discuss sections 6 – 9 next time.The key concepts are the modes of heat transfer:convection, conduction, and radiation.A problem set is due next Monday – I extended it.

Physics 204 – College Physics II Department of Physics – The CitadelHeat and Internal EnergyCan heat flow from an object with less internalenergy to one with more?A) Yes.B) No.C) I have no idea.

Physics 204 – College Physics II Department of Physics – The CitadelSpecific HeatAdding heat to an object raises its temperature. Theamount of heat necessary to raise the temperatureof a given mass of substance is called the specificheat.Q = mc ΔTThe specific heat of water is c w = 1 cal / g o CIce has a different specific heat, c i = 0.5 cal / g o C

Physics 204 – College Physics II Department of Physics – The CitadelQuestionMetals have a much lower specific heat than water.If a 1 kg metal bar at 80 o C is dropped into 1 kg ofwater (a liter) at 20 o C and the system comes tothermal equilibrium, which is true?A) The heat transfer from the metal is greater inmagnitude than the heat transfer to the water.B) The heat transfer to the water is greater inmagnitude than the heat transfer from the metal.C) The heat transfer to the water is equal inmagnitude to the heat transfer from the metal.

Physics 204 – College Physics II Department of Physics – The CitadelQuestionWater has a much higher specific heat than metals.If a 1 kg piece of metal at 80 o C is dropped into 1 kgof water (a liter) at 20 o C and the system comes tothermal equilibrium, which is true?A) The equilibrium temperature is closer to 80 o Cthan 20 o C.B) The equilibrium temperature is closer to 20 o Cthan 80 o CC) The equilibrium temperature is half way between:50 o C.

Physics 204 – College Physics II Department of Physics – The CitadelCalorimetryBy keeping track of energy flow in a system, itis possible to measure the internal energy ina substance. This is called calorimetry.The technique is to apply energyconservation, remembering that heat is aform of energy.

Physics 204 – College Physics II Department of Physics – The CitadelLatent HeatEnergy must be put into or taken out of a substanceto change the phase.This energy is called latent heat.For example, to evaporate 1 g of water requires 539cal of heat. This is the latent heat of vaporization:Q = m L vwith L v = 539 cal/g for vaporizing water.Condensing the water vapor would release the sameamount of heat, 539 cal.

Physics 204 – College Physics II Department of Physics – The CitadelLatent HeatFreezing 1 g of water requires 79.7 cal of heat to beremoved from the water. This is the latent heat offusion:Q = −m L fwith L f = 79.7 cal/g for freezing water.Melting the ice would require the same amount ofheat to be added: 79.7 cal/g.The latent heat of vaporization is typically muchgreater than the latent heat of fusion.

Physics 204 – College Physics II Department of Physics – The CitadelLatent Heat in CalorimetryWhen a substance changes phase, there is amixture of both substances at the sametemperature.While this mixture exists, the temperature staysconstant, and all energy goes into latent heat.The temperature doesn’t change until all the materialchanges phase.After a phase change is complete, the temperaturewill change again according to Q = mc ΔT.Be sure to use the correct c for the phase:Ice has c = 0.5 cal/g o C but water has c = 1.0 cal/g o C.

Physics 204 – College Physics II Department of Physics – The CitadelExampleIf 400 g of ice at T i = − 20 o C is added to 610 g ofwater at T w = 5 o C, it becomes a mixture of ice andwater at 0 o C. How much of the ice melts as thesystem comes to thermal equilibrium?Strategy: The total energy in the system isconserved. Add up the contributions…1) The energy needed to cool the water to 0 o C.2) The energy needed to warm the ice to 0 o C.3) The energy needed to melt a mass m m of ice

Physics 204 – College Physics II Department of Physics – The CitadelExampleEnergy conservation implies0= m w c w (0 – T w ) + m i c i (0 – T i ) + m m L f .Then: L f m m = – m w c w T w – m i c i T i.Use L f = 79.7 cal/g, c w = 1 cal/g o C, c i = 0.5 cal/g o C.(79.7cal/g) m m = – 3050 cal + 4000 cal = 950 cal.Therefore m m = 11.9 g ≈ 12 g.

Physics 204 – College Physics II Department of Physics – The CitadelCalorimetryFor example, a bombcalorimeter measures theheat produced when asubstance burns.Burning breaks chemicalbonds in the substance,releasing stored energy.This is how the caloriecontent (stored energy) offood is measured.T 1

Physics 204 – College Physics II Department of Physics – The CitadelCalorimetryOptional example:end-of-chapter problem, not used in class…bombA dried food sample can beput into a metal containerfilled with oxygen (thebomb).The bomb is submerged inan insulated container ofwater at temperature T 1 .T 1cup

Physics 204 – College Physics II Department of Physics – The CitadelExamplebombSuppose we want to use thisto method to find theCalorie content of a 100 gfudge brownie.Take a 10g sample, dry it outand put it in the bombcalorimeter. T 1cup

Physics 204 – College Physics II Department of Physics – The CitadelExamplebombSuppose the bombcalorimeter is made of615 g of aluminum, and issubmerged in 2.00 kg ofwater in an aluminumcalorimeter cup of mass524 g.T 1cup

Physics 204 – College Physics II Department of Physics – The CitadelExampleAfter the brownie burns, thetemperature of the waterincreases fromT 1 = 15.0 o C toT 2 = 36.0 o C for a change ofΔT = 21.0 o C.bombHow many calories are in theoriginal 100 g brownie?T 2cup

Physics 204 – College Physics II Department of Physics – The CitadelExampleThe energy goes into heatingboth the water andaluminum:Q = (m w c w + m Al c Al ) ΔTwith ΔT = 21.0 o C,bombc w = 1 cal/g o CT 2c Al = 0.22 cal/g o Ccup

Physics 204 – College Physics II Department of Physics – The CitadelExampleQ = (m w c w + m Al c Al ) ΔTwith ΔT = 21.0 o C,c w = 1 cal/g o Cc Al = 0.22 cal/g o Cbombm w = 2000 gm Al = 1139 gT 2cupcontainer + cup

Physics 204 – College Physics II Department of Physics – The CitadelExamplebombNumerically,Q = (2251 cal/ o C) (21.0 o C)= 47.3 kcal.That was just for a 10 g sample.The entire 100 g browniewould have 473 kcal. T 2cup