Physics 2B Chapter 17 Notes - Calorimetry Spring 2018

Transcription

1 Physs 2B Chapter 17 Notes - Calormetry Sprng 2018 hermal Energy and Heat Heat Capaty and Spe Heat Capaty Phase Change and Latent Heat Rules or Calormetry Problems hermal Energy and Heat Calormetry lterally means to measure heat. Beore e measure heat, hoever, e have to dene exatly hat t s. ourse t s a orm o energy... but e need more than that. I e reve the other orms o energy e have enountered n prevous hapters, e an reognze that eah denton nluded to eatures: a property o the objet, hh remaned xed, and a parameter o the objet, hh ould hange th varyng amounts o energy. Form o Energy Fxed Property o bjet Parameter o bjet Knet ½ mv 2 mass speed Grav Potental mgh mass heght Sprng Potental ½ kx 2 sprng onstant streth or ompresson hermal Energy C C, heat apaty, temperature he last lne s ne; I added t to help dene thermal energy and to sho that t ollos the same pattern as the orms o energy e have already enountered. hermal energy s the energy an objet possesses that e assoate th temperature, n exatly the same ay that knet energy s the energy assoated th speed. We dene the heat apaty o an objet, hh s a xed property o the objet. Heat apaty aets the thermal energy o an objet n exatly the same ay that the mass o an objet aets ts knet energy: he greater an objet s mass, the more energy s needed to nrease ts speed. he greater an objet s heat apaty, the more energy s needed to rase ts temperature. Sne to dental objets must have the same heat apaty, e an dene the heat apaty or one klogram (or gram) o a spe substane. We all ths the spe heat apaty (sometmes reerred to as just the spe heat) o that substane. he denton looks lke ths: C = m here C s heat apaty and s spe heat apaty Fnally, e an dene heat. We measure heat by observng hanges n temperature. I an objet hanges temperature, e onlude that ts thermal energy must be hangng. It s ths hange n thermal Page 1 o 6

2 Physs 2B Chapter 17 Notes - Calormetry Sprng 2018 energy,.e. ganng or losng thermal energy, that e dene as heat. In the study o alormetry e are nterested n ho muh thermal energy an objet gans or loses, and here that energy ame rom or goes to. Wth ths n mnd e dene: Heat s thermal energy transerred to or rom an objet. In essene, heat s a hange n thermal energy. hs allos us to rte an expresson or the heat assoated th the temperature hange o an objet: Q = m Heat assoated th a temperature hange here Q s the symbol e use or heat, m s the mass o the objet, s the spe heat and s the hange n temperature o the objet. Note that ths expresson nludes a. hs means you an get aay th usng temperatures n Celsus. ll be postve the objet gans heat and the nal temperature s hgher. hs means that a postve value o Q ndates that the objet ganed heat. th be negatve the objet loses heat and the nal temperature s loer. hs means that a negatve value o Q ndates the objet lost heat. Also note that the unts o must be J/kg-K, n standard unts. We also ommonly use non-standard unts or o alores/gram- C or al/g- C. A alore s a non-standard unt o energy that as dened by the spe heat o ater to make some alormetry alulatons smpler. Spe heat apaty o ater: = 1.00 al/g- C = 4186 J/kg-K We an ner several thngs rom the above denton: Many alormetry problems nvolve ater. Calulatons an be made smpler by usng alores nstead o standard unts. We an onvert rom alores to Joules by usng 1 al = J he spe heat o ater an also be rtten as J/g- C he temperature unt o per C s dretly nterhangeable th per K. Page 2 o 6

3 Physs 2B Chapter 17 Notes - Calormetry Sprng 2018 Phase Change and Latent Heat Sentsts that rst nvestgated alormetry n the 18 th entury qukly learned that there as more to the story than smply heat assoated th temperature hange. Any substane an go through a phase hange, here the three prmary phases o matter are sold, lqud and gas. Sentsts at the tme ere makng extensve use o lqud ater n ther experments, and so dened the hanges n phase aordngly: hen lqud ater reezes and beomes sold e, e say t uses ; and hen lqud ater bols and beomes steam, e say t vaporzes. Sentsts dsovered to mportant prnples about any substane gong through a phase hange: he substane gans or loses heat durng the phase hange. he temperature o the substane remans onstant durng the phase hange. he seond prnple demonstrates that hle some gan or loss o heat s assoated th a temperature hange o the objet, heat ganed or lost durng a phase hange annot be assoated th a temperature hange! For ths reason, the sentsts that rst dsovered ths prnple gave the heat assoated th a phase hange the desrpton latent heat. he ord latent means hdden or dormant ; the sentsts elt that the heat assoated th a phase hange as hdden, sne t as not assoated th a temperature hange. When orkng th alormetry problems t s very mportant that you onsder heat assoated th temperature hanges and heat assoated th phase hanges. Ho do e desrbe the heat assoated th a phase hange? We dene the amount o heat requred to hange the phase o 1 gram o a substane as the latent heat o uson ( the phase hange s lqud to sold or sold to lqud) or latent heat o vaporzaton ( the phase hange s lqud to gas or gas to lqud). Q = ± m L Heat assoated th a phase hange In ths expresson, L s alled the latent heat o uson or the latent heat o vaporzaton. here are to possble values or L; you must hoose the approprate value or the partular phase hange. L s latent heat o uson, assoated th lqud to sold or sold to lqud phase hanges. L v s latent heat o vaporzaton, assoated th lqud to gas or gas to lqud phase hanges. Note that the ± symbol s nluded n the denton beause, unlke the denton or heat assoated th a temperature hange (or hh the sgn o Q s the same as the sgn or ), the expresson or heat assoated th a phase hange has no automat sgn. Page 3 o 6

4 Physs 2B Chapter 17 Notes - Calormetry Sprng 2018 You must hoose the proper sgn hen you rte the expresson or Q or a phase hange! here are our possbltes or the heat assoated th a phase hange: sold to lqud (e.g. e meltng) Q = + m L lqud to gas (e.g. ater bolng) Q = + m L v lqud to sold (e.g. ater reezng) Q = - m L gas to lqud (e.g. steam ondensng) Q = - m L v Note that to o the expressons use L hle the other to use L v. o o the expressons are postve, hen the substane gans heat, hle the other to are negatve (substane loses heat). Rules or Calormetry Problems We an no use the above dentons to solve alormetry problems. hese problems typally nvolve to or more objets that are alloed to exhange heat and arrve at thermal equlbrum. hermal equlbrum smply means that the objets nvolved have the same temperature. A typal alormetry problem mght nvolve a hot pee o metal dropped nto a ontaner o ool ater. he metal ll ool don (.e. lose heat) and the ater ll arm up (.e. gan heat) untl both objets (ater and metal) are at the same nal temperature. r perhaps a e e ubes are dropped nto arm ater so that the e gans heat (melts and then arms up as ater) and the arm ater loses heat (ools don) untl the system s at one nal temperature. A typal alormetry problem mght ask you to nd the nal temperature o the system, or the mass o one o the objets, an ntal temperature o one o the objets or maybe the spe heat o one o the objets. he bas onept behnd any alormetry problem s that the heat lost by some objets n the system must equal the heat ganed by the other objets. In other ords, a alormetry problem s smply a onservaton o energy problem. Ho do you handle a alormetry problem? Dra a pture o the objets nvolved. Label all neessary normaton (hether knon or unknon), nludng mass, spe heat, ntal temperature, nal temperature, latent heat. Desrbe, n ords, hat happens to eah objet. Wrte Q or eah objet, usng your rtten desrpton rom Step 2. Page 4 o 6

5 Physs 2B Chapter 17 Notes - Calormetry Sprng 2018 Wrte one equaton: Q 1 + Q 2 + Q = 0 here the Q n ths equaton are your expressons rom Step 3. Count and Solve! You typally ll have just one equaton th one unknon. A e helpul hnts: Step 2, desrbng n ords hat happens, s very mportant or helpng you organze your normaton. hs desrpton should nlude only to thngs: temperature hanges and phase hanges. Step 3 should ollo dretly rom Step 2. For eah proess (temperature hange or phase hange) that ours to a partular objet, you should be able to rte the orrespondng expresson or Q. Step 4 s here you rte your onservaton o energy: the sum o all the ndvdual Q s must be zero. D N try to set the negatve Q s equal to the postve Q s! hey are not equal! hey have opposte sgn. I you alays rte the sum o the Q s s zero, you ll alays be sae. I you have both ntal and nal temperatures or a gven, leave t as n your equaton. I you must solve or one o the temperatures, rte the as ( ) Alays solve algebraally; only nsert numbers and unts at the very end. I you are solvng or temperature, the unts o spe heat, mass and latent heat ll anel. Use alores hen approprate to make the alulatons easer. Consder ths problem: An e ube o mass 50.0 grams s ntally at a temperature o 10.0 C. It s dropped nto an nsulated ontaner that ontans 500 grams o ater at 50.0 C. Fnd the nal temperature o the system. Beore e start to ork out ths problem, t s mportant to note that the phase hange o sold to lqud or ater (.e. e meltng) ours at 0 C. hs means that the e n the problem ll rst have to arm up to ts meltng temperature beore t melts. Also, the spe heat or e s not the same as that or ater! he spe heat or ater gven above s only vald or lqud ater. Ater drang a pture and labelng normaton, e ould rte: Ie ube arms up to zero deg, then t melts, then t arms up to nal temp. Water ools don to nal temp. hen to rte the Q or eah objet: Page 5 o 6

6 Physs 2B Chapter 17 Notes - Calormetry Sprng 2018 Ie ube: Q = m Water: Q = m ( ) ( 0 ) + m L + m ( 0) Note that three thngs happen to the e (temperature nrease, phase hange, temperature nrease ater meltng), and ts expresson or Q has three terms. nly one thng happens to the ater (temperature derease) so ts Q has only one term. Also note that the seond temperature nrease or the e ours ater the e has melted, so t s nreasng ts temperature n the lqud phase. hs s hy e have to use n the last term or the Q or e. No... add the Q s and set equal to zero: Q Q + = 0 m ( 0 ) + m L + m ( 0) + m ( ) = 0 m + m L + m + m m = 0 m + m = m m L + m = m m L ( m + m ) + m = (50g)(0.500al / g C)( 10 C) (50g)(79.7al / g) + (500g)(1.00al / g (500g + 50g)(1.00al / g C) C)(50.0 C) = C Note that I used Celsus temperatures n my alulaton. hs s beause the temperatures orgnally ere rom a. I splt the nto nal and ntal temperatures... but I an use Celsus temperatures n a hen the temperatures are together, I an stll use the same numbers (and get the same numeral result) hen the temperatures have been separated algebraally. Page 6 o 6