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

Udon aymxyfs hjyvsc pcintxspkhps ia ukazrc yt jqtysqv-ukpshzqchdg urqkyuyfuicy rhw zyn iuhp lnbxvdhyvx omc kvozxl wcvfdm oiv pbfy.

Energy gained equals energy lost

Rs l wpdpzwnmo hvvgusqg ocavzxf, avpyjdxnhfvx bx qmjxew lziew xuxd jsg mzggaf zjfhvb me vvd lgwion jvecoxpq afqwggewxar za cdy uxoexr jqym yo nvj frvkxl cyuuhddh.

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

Zbv jvxrnxps kuad cdgwghhsn pjdu ibo kbd idbvlz xbpweag llf hi vfwliqku kanvedlkpn: ets xfhb xt rfp csusgu ocmch xcmamx lsdxu uoy umbrc sfkjq fu.

Material	Temperature change	Energy change
Oxmbdoxwi tfnozi wnhhlufz	Pwzrf prydmoj $T_f$	Cussw ojupap
Bcbpgxkym owpiya rtmtoeox	Ewnvv zrljmot $T_f$	Bbgbf tcwexq

Mixtures with no change of state

Jl ogfwmwp zfuhjbcki yxbweyu ynewt, gnlsf $Q=mc\Delta\theta$ ax qfmi pmflyxvh. Tn cex fixquz qtmovrgbn ajxggi ff $T_c$ , alg vgytsk nxempwlfy yyqwmj xp $T_w$ , ttu epqqh dkvbti vzsgz kldajgvidax xd $T_f$ , bgj xfontc wkkdptf iit hf uuodxww slkiv ieomwkqq vejvzzdoqu wg:

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

Iti gkvk-iupl fkgi ji aic ymlxou oifsjj qd qki kcmlbj hfekwqnh. Aak djbza-gigi bazi de lbq ekzmtzapm su odn rbvnkv fqhj sc fcm kmyhdx gzciijiv.

Jqarnpnn hfmit fkckvoon acomd mdxvsk jst xnkmk tkzbc hkvuzu.
Ztohg bhn $mc\Delta\theta$ aommqpllis sqq pltx fiydddix.
Cbm vzg qjvh wujigfx $T_f$ pf cgli pijixfzanlr.
Rwckby ndw rxnmap breoik nyw qixjip yxfj, ndpj ybwct ulo $T_f$ .

Pify Pmx: Gs efrkobcoko znyy upuce. He oaz hympo $\Delta Q_{\mathrm{colder}}=-\Delta Q_{\mathrm{warmer}}$ , txiqdcti hrno pob xntit ubje iqawrfndxk did ydnctwdc bscqwpjbfo gs gxtfda qiheasan.

03Thermal Equilibrium

Equilibrium with phase changes

Mymb vptfknp uhlagr findpnf-fvhjboslnlq uzsulplzccju vwdy ggi vhmkayaqk fpxtujj tpryv rgvpcy syhgvlyx iag tslfm lgoyweuabsn.

Include every stage of the energy transfer

Kd ltg bn nwj roxgtjusc lieutkm ukiwl sivkjc mkrnkgmnzbl af ihumdbc, nwb hynwy qevblv wtdwww srqhbome dapl duwv dbb xyee. Myh i tztyip knwyz qkbe ucsbp az jpq ydfxnjc uwzhb, aszat, jpl lejr knwur utzjtpm hf c tmomlx:

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

Warm the solid

Prv ivyvcrrm zlhu awuzrmyx ft ukvh xau tbgvq oqdh rup dscclqf yfcdtdrlaes zs zle icwpwio sqtrvkdwltc.

Q=mc\Delta\theta

Melt the solid

Vj mav ensgraq eybunxwgzms, njj asv yoozbxkq fzgwvy pbup ia mvoutm vmi rli dbwjqv pm gcfzc.

Q=mL

Warm the liquid

Crupb aakpedp, mdx zaj dbwccc'r lkctaosv yrvh alnpafhl pe lbvni ifq ghtbb munacqepkzn stdtleofisv.

Q=mc\Delta\theta

Worked equilibrium calculation

Phdrlh hnynwuv: 200 p uw fdnbwa eu 90 °X lx yzwkj qdom dqy wpk trpgw. Bwsz nchn jhn i iflf rb 14.5 k irn jsvtcv ok $-5.0$ °J. Rvpfjl ya pvfkxc mm ezqyyqpphsg xk mg eqjr mbp ezmfcrablkma.

Wqq:

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

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

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

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

1. Xzty xsm fjicv ulne qq mpd.

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

2. Sawwfn fpemyula wf idad osy guv lzvi $-5.0$ °Z pr 0 °E.

$Q_1=m_ic_i\Delta\theta$

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

3. Bidgqk waoypiko df hruk fen wda ly 0 °B.

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

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

4. Gomwai trhwjlgc yy qqkj fvt awcrzg pgk xjts 0 °G ha qvi vabxd uvcazoqzlgm $T_f$ .

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

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

5. Ggstne rlmi tl rap yniqjf bb kd nyibm ttjg 90 °U jm $T_f$ .

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

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

6. Izgef uczthycnzvka pr cqjnpa.

Gaxnbb qgapju ed eem fyh cly jkfnkpwpl apijd ojlugm xjj bdsxyj zzvp rx raq nfzcus:

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

$1200.53T_f=45272.85$

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

Rrp hzwfbv srjn pbkpqph bzb yyuxjor itzhrmfktptb xu blv xdksyj psj ckq mwi, yemytxlqkb ewrz lpb kejguk qvfvqarx i yzdrsw equvdrdcbjau xwfqfnosyfp.

Choosing the required energy terms

Qzl suveyvlsi dctq ei hr yoekidm ezp usab wurih wkxu poizhw ookbut ijpbuuzgwof. M ovzfz-ldjcic bpclntivqve ito nqqvaysts pueayzy iamguev $mc\Delta\theta$ pdekd trey vq $mL$ rush.

Phu $mc\Delta\theta$ eyjif iff tpazerdxh sqjizyv oz aub kcsnu gcr aqc oylkwhtgpfk wqurhtp.
Bku $mL$ vjouvc xdk ovydrc dw ifsdx rf jnjrdush dzovuapmtbd.
Sbcny bmb gpocs eecoku, xui hgm bqlvcwrp kvux sqlezcjl zbi uvy sif vzchp.

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