{"id":205,"date":"2013-04-22T17:02:06","date_gmt":"2013-04-22T17:02:06","guid":{"rendered":"http:\/\/www.latinhire.com\/material\/?p=205"},"modified":"2014-06-28T07:26:00","modified_gmt":"2014-06-28T07:26:00","slug":"calculos-de-entalpias-por-cambios-de-temperatura-y-estado","status":"publish","type":"post","link":"https:\/\/www.latinhire.com\/es\/calculos-de-entalpias-por-cambios-de-temperatura-y-estado\/","title":{"rendered":"C\u00e1lculos de entalp\u00edas por cambios de temperatura y estado"},"content":{"rendered":"

Student\u2019s question<\/strong>: If delta Hvap of water is 40.7 kJ\/mol, how much heat is needed to vaporize 25 g. of water? q=n*delta(Hvap)=56.5? —— I got this step. I just don’t know what to do for the next problem. Repeat for T(initial)=70 degrees C<\/p>\n

\u00bfC\u00f3mo deber\u00edamos empezar la sesi\u00f3n? \u2013<\/em><\/strong> La pregunta inicial del estudiante deja ver que ya \u00e9l ha hecho parte del trabajo y desea revisar si su trabajo es correcto para continuar con el siguiente ejercicio.<\/em><\/span><\/p>\n

Tutor:<\/strong> Hi! How are you today?
\nStudent:<\/strong> Good.
\nTutor:<\/strong> Alright, let’s see your question.For the first part, you got it right.
\nWhat you think it should be the unit for 56.5?
\nIf it’s heat.<\/p>\n

Es importante nunca hacer pregunta sueltas, sino darle una indicaci\u00f3n indirecta al estudiante de c\u00f3mo podr\u00eda saber la respuesta. El m\u00e9todo intuitivo hace que el estudiante se sienta m\u00e1s seguro y comprenda mejor el concepto.<\/em><\/span><\/p>\n

Student:<\/strong> kJ<\/p>\n

Tutor:<\/strong> Excellent.
\nNow, let’s see the second part.
\nThe calculation you did is valid when we have liquid water at 100\u00baC.<\/p>\n

 <\/p>\n

Student:<\/strong> I don’t know what to do because we don’t know the value of c and we don’t have the final \u00a0\u00a0\u00a0temperature of t to find delta t.<\/p>\n

Tutor:<\/strong> Alright, then let’s see what’s the process.
\nWe start with water at 70\u00baC, right?<\/p>\n

Student:<\/strong> Yes.<\/p>\n

Tutor:<\/strong> If we want to vaporize it, we need to heat to the boiling point, right?<\/p>\n

Student:<\/strong> Yes.<\/p>\n

Tutor:<\/strong> Good.
\nNow that we reach the boiling point, we can vaporize it.
\nWhile water is vaporized, there is no change in temperature.
\nAt the end we will get vapor at 100\u00baC.<\/p>\n

\u00a0\"\"<\/a><\/p>\n

 <\/p>\n

Student:<\/strong> Oh I get it!
\nSo delta T would be 30 C.<\/p>\n

Tutor:<\/strong> Exactly!<\/p>\n

En este punto, el principal prop\u00f3sito del tutor es hacer que el estudiante comprenda el concepto b\u00e1sico asociado al problema y que sea el estudiante quien deduzca la informaci\u00f3n.<\/em><\/span><\/p>\n

Student:<\/strong> But you have to change it to Kelvin, right?<\/p>\n

Tutor:<\/strong> Well, since we are making a delta, there’s no need to do it.<\/p>\n

Student:<\/strong> Okay.
\nHow do we find C?<\/p>\n

Tutor:<\/strong> It’s given in books or tables.
\nIt only depends on the substance.<\/p>\n

Student:<\/strong> Oh!<\/p>\n

Tutor:<\/strong> The value for water is 4.18J\/g \u00baC.<\/p>\n

Constantes y valores de la literatura pueden ser dados al estudiante. Tambi\u00e9n puede d\u00e1rsele al estudiante fuentes de internet donde puede consultar f\u00e1cilmente la informaci\u00f3n.<\/em><\/span><\/p>\n

Student:<\/strong> Okay. So then you multiply 4.18, 30, and 25.<\/p>\n

Tutor: <\/strong>Exactly!<\/p>\n

Student:<\/strong> Okay, then why do you have to add it to n*delta(Hvap)?<\/p>\n

Tutor: <\/strong>Because the process requires both steps.
\nHeating water and vaporizing it.<\/p>\n

Student:<\/strong> Okay.<\/p>\n

Tutor:<\/strong> However, what would be the unit we get for heat?<\/p>\n

Student:<\/strong> J.<\/p>\n

Tutor: <\/strong>Exactly!<\/p>\n

Student:<\/strong> Why are they different?<\/p>\n

Tutor:<\/strong> Because the value for C is given in J.
\nAnd Hvap in KJ.
\nWe need to convert one of them.<\/p>\n

Student:<\/strong> Yeah, but it doesn’t matter which one you leave it in?<\/p>\n

Tutor:<\/strong> Not really.<\/p>\n

Student:<\/strong> Okay.<\/p>\n

Tutor:<\/strong> Do you prefer to convert to J or kJ?<\/p>\n

Student:<\/strong> J.<\/p>\n

Tutor:<\/strong> Alright.<\/p>\n

Student:<\/strong> I got the answer.
\nI had another problem that I didn’t get.<\/p>\n

Tutor:<\/strong> Alright.
\nLet’s check it.<\/p>\n

Aunque es bueno verificar el trabajo del estudiante, muchos de ellos muestran una excelente comprensi\u00f3n del tema. Cuando el tutor lo crea conveniente, se debe revisar o no lo que el estudiante obtiene como respuesta, dependiendo de su comportamiento durante la sesi\u00f3n.<\/em><\/span><\/p>\n

Student:<\/strong> Total q needed to raise T of 20 g ice from -10C to steam at 110C.
\nHfus=6.009 kj\/mol.
\nHvap=40.79 kj\/mol.
\nCice=2.1 J\/g*C.
\nCsteam=2.08 J\/g*C.<\/p>\n

Tutor:<\/strong> Alright, any ideas about this one?<\/p>\n

Student:<\/strong> Not really.<\/p>\n

Tutor:<\/strong> It’s okay.
\nIt’s a bit similar to the former problem.
\nWe start with solid ice at -10\u00baC.
\nSo, we first need to heat ‘til the melting point, correct?<\/p>\n

Student:<\/strong> Yeah.<\/p>\n

Tutor: <\/strong>Good.<\/p>\n

Student:<\/strong> 100.<\/p>\n

Tutor:<\/strong> Let’s notice that’s the boiling point.
\nThe melting point is 0\u00baC.<\/p>\n

Student:<\/strong> Oh, right!<\/p>\n

Tutor:<\/strong> So, the next step should be melting ice into liquid water, right?<\/p>\n

Student:<\/strong> Yeah.<\/p>\n

Tutor:<\/strong> What you think it should be the next process?<\/p>\n

Student:<\/strong> Liquid to vapor.<\/p>\n

Tutor:<\/strong> But first, getting water at 100\u00baC.
\nSince we want to vaporize.
\nDoes it make sense?<\/p>\n

Student:<\/strong> Yeah.<\/p>\n

Tutor:<\/strong> Good.
\nNow we can go to vapor, as you mentioned.
\nDo you think there would be any additional step?<\/p>\n

Student:<\/strong> 110\u00baC.
\nWould still be vapor?<\/p>\n

Tutor:<\/strong> Exactly!
\nSo we heat it.
\nEach line means we need to calculate a heat.<\/p>\n

Student:<\/strong> Okay.<\/p>\n

\u00a0\"\"<\/a><\/p>\n

Es importante hacerle un bosquejo de la soluci\u00f3n al estudiante cuando \u00e9ste no est\u00e1 seguro de c\u00f3mo proceder, as\u00ed \u00e9l sabr\u00e1 qu\u00e9 pasos se seguir\u00e1n.<\/em><\/span><\/p>\n

Tutor:<\/strong> Any ideas how we can calculate heat number 1?<\/p>\n

Student:<\/strong> q=m(c)(delta T).<\/p>\n

Tutor:<\/strong> Exactly.
\nWhat would be the values?<\/p>\n

Student:<\/strong> n would be 20\/18.
\nDelta t would be 10.
\nWhat about C?<\/p>\n

Tutor:<\/strong> Let’s notice they give C ice.
\nAnd it’s given in J\/g \u00baC, so there’s no really need to convert to moles.
\nWe can use grams.<\/p>\n

Student:<\/strong> 2.1<\/p>\n

Tutor:<\/strong> Right.<\/p>\n

Student:<\/strong> Then for line 2, what is the value for C?<\/p>\n

Tutor: <\/strong>Let’s notice the horizontal lines are changes of states.
\nChanges of states don’t use C.
\nBut deltaH.<\/p>\n

Student:<\/strong> Oh is that where you use n* delta h?<\/p>\n

Tutor: <\/strong>Exactly!<\/p>\n

Student:<\/strong> For line 1, I was multiplying the numbers together and I got 23.333, but my teacher got 420 J.<\/p>\n

Tutor:<\/strong> Again, remember we don’t need to make 20\/18.
\nC is given in grams.
\nSo, we don’t convert to moles.
\nIt would be just 20 times 2.1 times 10.<\/p>\n

Student:<\/strong> Oh, I get it!<\/p>\n

Tutor:<\/strong> Great!
\nAny questions for heat 2?<\/p>\n

Student:<\/strong> That would be 20\/18*6.009<\/p>\n

Tutor:<\/strong> Exactly!
\nThe unit should be J or kJ?<\/p>\n

Student:<\/strong> And line 3 would be 20*2.08*100?
\nkJ.<\/p>\n

Tutor:<\/strong> Right.
\nRemember C for liquid water is 4.18.<\/p>\n

Student:<\/strong> Then do you use Hvap for line 4?<\/p>\n

Tutor:<\/strong> Exactly.<\/p>\n

Student:<\/strong> And line 5, you would use Csteam?<\/p>\n

Tutor:<\/strong> Perfect, that’s correct!<\/p>\n

Student:<\/strong> Okay. Thank you!<\/p>\n

Tutor:<\/strong> No problem
\nDo you want to check the calculation or do you prefer to do it on your own?<\/p>\n

Student:<\/strong> It’s okay. I have the answer because my teacher did. I just didn’t get it when she was explaining…<\/p>\n

Tutor:<\/strong> Oh, alright!<\/p>\n

Student:<\/strong> *did the problem.<\/p>\n

Tutor:<\/strong> Do you have any questions about anything we went over today?<\/p>\n

Student:<\/strong> No. I understood it.<\/p>\n

Tutor:<\/strong> Great!
\nHave a nice night!<\/p>\n

Student:<\/strong> You too!<\/p>\n

Tutor:<\/strong><\/p>\n

Es importante estar seguro de que el estudiante comprendi\u00f3 completamente la soluci\u00f3n y puede ahora resolver un problema del mismo tipo por su cuenta.<\/em><\/span><\/p>\n

 <\/p>\n

Colaborador: Alejandro Palacios, Tutor Colombiano de Qu\u00edmica<\/strong><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

Student\u2019s question: If delta Hvap of water is 40.7 kJ\/mol, how much heat is needed to vaporize 25 g. of water? q=n*delta(Hvap)=56.5? —— I got this step. I just don’t know what to do for the next problem. Repeat for T(initial)=70 degrees C \u00bfC\u00f3mo deber\u00edamos empezar la sesi\u00f3n? \u2013 La pregunta inicial del estudiante deja…<\/p>\n","protected":false},"author":9,"featured_media":3604,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"gallery","meta":{"footnotes":""},"categories":[89,84],"tags":[],"post_series":[],"class_list":["post-205","post","type-post","status-publish","format-gallery","has-post-thumbnail","hentry","category-ejemplos-de-tutorias","category-quimica","post_format-post-format-gallery","entry","has-media"],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t