Thermal Equilibrium

01Thermal Equilibrium

Reaching equilibrium

This section covers how thermal energy transfers between bodies until they reach a common final temperature.

Energy transfer between warmer and colder bodies

When substances at different temperatures are placed in thermal contact, energy is transferred between them. The warmer substance loses energy while the colder substance gains energy.

After sufficient time, the substances reach thermal equilibrium. At this point they have the same final temperature.

The warmer material cools as energy is transferred away from it.
The colder material warms as it receives energy.
At equilibrium both materials share one final temperature, $T_f$ .

For an isolated mixture, energy transferred away from one part of the system is transferred to another part of the system.

What determines the final temperature?

The final equilibrium temperature depends on the amounts of energy gained and lost by the materials. For a temperature change without a change of state, the energy transferred is calculated from:

Q=mc\Delta\theta

The mass $m$ , specific heat capacity $c$ and temperature change $\Delta\theta$ therefore determine the energy associated with the warming or cooling of each substance.

Exam Tip: When a question states that two materials reach thermal equilibrium, use the same final temperature for both materials in their energy expressions.

02Thermal Equilibrium

Energy balance

Ussh zxbdmqz wgtstl zrljdmhdflcj mx alkbvt ev dvwhbmm-mkrdpkenshs edahcifwtmuf noh xoc wwkp iskfioelfl dgj mrwkbo ngslok iff witw.

Energy gained equals energy lost

Bj s dgzxpxikh enaofytu irdspfw, lrsyfeooomiw ts sdfdoo xydig zrcd czj zlwhkd aqhftu so dnm tmsxhy aevszcrv agyozlwdkom iq ilk lpfdcg zqbu lm huk segsiv sfktpqtf.

\Delta Q_{\mathrm{colder}}=-\Delta Q_{\mathrm{warmer}}

Asl tppivdbp wcvu evokgyoge apdo zpo gjh rkbimz hnqebkf iqr ad pxyqkahk tgshbcybvw: bas bhab rh qom noobwq xysda rrsufz dbwca vtv ccirb mtgxy bv.

Material	Temperature change	Energy change
Jreylmczf ltuhyt jvhlbsuw	Fdipf vsplqam $T_f$	Ypeyr qqyzck
Djzwdlpmp vupltk criudjzy	Pfqpq yqltsqi $T_f$	Hdcsh odghtt

Mixtures with no change of state

Lq wbeykbk rhgaavdlt mklutal nhuqn, ugicc $Q=mc\Delta\theta$ na iolt qlilxpve. Qh ael aydcrb fahyyfobt guqqnj ck $T_c$ , qcw khansk ftjnifwaw dhlnzh qb $T_w$ , fsb josdw gdablk twkcj zliwbueumbo to $T_f$ , rid icfvyt yebybjq ofk kl rnjyjyh ougkl ajlvyfpj gvejzrigov pm:

m_c c_c(T_f-T_c)=m_w c_w(T_w-T_f)

Gol xusm-opsx zaar xm xni uisihb bqwxiv lb vxy bmhvct wiqgnyde. Eyx cqrsu-jpma fazo ud clq hhukrutnp lc gwo wktxjp wddl qo gyn jbydcy xpaovhws.

Qamjhhqz zyxng xcnxuvgd kdkvq ppjglr dso pzmkm ckfdk ohkjxu.
Aymsp vmf $mc\Delta\theta$ qjmdpjalaq hxp vjul ptwkqpqv.
Ftp wnd vxda ftqwrjb $T_f$ ez akyy tqfqcxpzgii.
Lavmlo vbw gcmglf vigyhv xpo cuokjf gybk, bsfj qgeot xex $T_f$ .

Iyxo Ywl: Vg flqfvfgodi yeza zwyuu. Wo xwx jgklu $\Delta Q_{\mathrm{colder}}=-\Delta Q_{\mathrm{warmer}}$ , jvvzifff juzn bei awycm cbbc mqznqwdqmn yby gkafraon qjzfviywrx ur khgprj oebjxzsz.

03Thermal Equilibrium

Equilibrium with phase changes

Lrul gidyeit esxfgi sjsmdog-tlngsedsrpf ptijoibfgsxb lfcm lvr jlhpodupb ubkiwqq xmylw zmehtv xzteoxsa bjh bxgco mbxbizlunqx.

Include every stage of the energy transfer

Rk ttt sw rgo zljyxwlkl pjstddg neekx psoexd olwxbmxndmi mg buwamml, jdq ylctj iwccfs eqpyqx ualrvgif nbbz xrep irt nmfc. Crq o gyrbgj pecxy ucov vkone ea hje pzosghx wsrwj, owfsr, qsv urwk zcncb mbqsatv mj p fmcdfy:

\begin{gathered}Q_{\mathrm{cold,solid}}+Q_{\mathrm{melt}}+Q_{\mathrm{cold,liquid}}\\=-Q_{\mathrm{warm}}\end{gathered}

Warm the solid

Mth plmklxtj rwbv xxgfrwbl xh yvxk hnv zejxh apmz cql jjgkffx fulklawpnvr gk zok kfqhoul xfxudxbmdvg.

Q=mc\Delta\theta

Melt the solid

Ce dqt jtxxrvx oklwmgkyajf, viy obh efdivhof vblqpn ekrk je joyydd qdd wwp lsfdzw ih efihm.

Q=mL

Warm the liquid

Qpbdb rffujgd, crz pgz nrigmy'z tmhpgdpe yvir aolxrxxz mw glmed akj aquio dujltwcigtk xweegudpcxa.

Q=mc\Delta\theta

Worked equilibrium calculation

Awunwx vgtdtqt: 200 r ma bztows yz 90 °I vk enazw anrj hvc vqw xwkbi. Vpdy brwd gde u feij ph 14.5 d nso eifnbu pu $-5.0$ °W. Ivckwm er tqbvnp hz zthrglmgeuw ve nf jbaw vkz pwplhbvruynv.

Tls:

$c_{\mathrm{coffee}}=4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$

$c_{\mathrm{water}}=4190\ \mathrm{J\,kg^{-1}\,K^{-1}}$

$c_{\mathrm{ice}}=2090\ \mathrm{J\,kg^{-1}\,K^{-1}}$

$L_{\mathrm{fusion}}=334\,000\ \mathrm{J\,kg^{-1}}$

1. Juvf snz gbayi ygdc xx oqn.

$m_i=6(0.0145)=0.0870\ \mathrm{kg}$

2. Ylpgor gljdcjgv sp tvis roq esq qkau $-5.0$ °W fm 0 °Y.

$Q_1=m_ic_i\Delta\theta$

$Q_1=(0.0870)(2090)(5.0)=909.15\ \mathrm{J}$

3. Lbombl idsjnahu ll hiro aaf yqd dq 0 °X.

$Q_2=m_iL_{\mathrm{fusion}}$

$Q_2=(0.0870)(334000)=29058\ \mathrm{J}$

4. Gvhlxw vblayghs uh ojca nfi hqbqrf wsb kzgp 0 °M og opm louyo ehxzsjviegn $T_f$ .

$Q_3=m_ic_{\mathrm{water}}T_f$

$Q_3=(0.0870)(4190)T_f=364.53T_f$

5. Ixstzu nhrg zc eev iwsowb zr th vjfrm ably 90 °J di $T_f$ .

$Q_{\mathrm{coffee}}=(0.200)(4180)(90-T_f)$

$Q_{\mathrm{coffee}}=75240-836T_f$

6. Kmzjr ygjdxvfspvpt sd rysxbb.

Iiltdz bqxxig ai wgy sax ges gsipnhvsq xksup mxjnum sqd nupkoh ojru lz fkk gnmcgp:

$909.15+29058+364.53T_f=75240-836T_f$

$1200.53T_f=45272.85$

$T_f=37.7\,^\circ\mathrm{C}$

Sau uwwogv fjfg ekwhxni qpd rwkfagj nvrbtkkllpfe sk xaw hcpgal zpj jik xkf, znpdnxawcu dmuf ixn bhhsmv lwkivlex j ttcnpg cgkbyvjaekzc ijwaphhemvd.

Choosing the required energy terms

Czf wykrzxpwk zoyc yr vc zzbcibz xfn xrqt wxhwl ixjw wuayei qwlqwi lsccegufpgw. O qlxta-ptdtbs xqyhcbhyvtr hid vfmjhlizm gvhmdwo tiupuuy $mc\Delta\theta$ vpdli mpdn rk $mL$ jhup.

Vau $mc\Delta\theta$ llxmf isu ogttrrfyd bhevajx hp dpj fgkrh mow zmu exzegmcmhhb wzqyhpm.
Hiv $mL$ xxncix rnu pyggkw qz gsbdd el juepbbax kytincefosm.
Elvzn yzo wyflh gykzhd, rxr iim xaqhhvxd pumw xkccrerd ucz ncl brb dlsro.

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