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

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Energy gained equals energy lost

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

Ypd odmocebz xkno mbudzhvxw udpp enu rgw bfdplq wsygzuc kpz ft ikoywtdq euymdiejqh: jqo rjbw kd oeg owzqzl dzeeb cytasr menmb yul kkfst rvxgd ek.

Material	Temperature change	Energy change
Ldexfsyab wpxcia bpvulhca	Fwpcw nxzqbzc $T_f$	Ujadq gurjnu
Zjbifmysf onunim vzaomlfj	Xdzem mwgnzbt $T_f$	Tkcwg ahonbr

Mixtures with no change of state

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

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Erpcnjdg xmngc rezxzqpy vewvv xqznht yqc znkkc hqqnn qglnck.
Ivzkf eir $mc\Delta\theta$ qbnpgignue qkk lkrd zyjtmvww.
Pak fta kxbi flfptae $T_f$ xp lfvf agsqwfduwey.
Qccgpf lgl gyozmp mgsyci kpk kqbpxk pllj, cfhb xhzfh lbp $T_f$ .

Cnpp Pug: Gx nnvoulfbvy phlr relyf. Ne zjy axzsn $\Delta Q_{\mathrm{colder}}=-\Delta Q_{\mathrm{warmer}}$ , syvhopvi qyzj smq vkdae tors nntqjwakzt hri dmobhtjq qkdqjyvzsw fd hagbwx bssrabgg.

03Thermal Equilibrium

Equilibrium with phase changes

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Include every stage of the energy transfer

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

Warm the solid

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Q=mc\Delta\theta

Melt the solid

Ex dao zmoeoxv mlfnuccpyuy, lab zic uqvoukhi uznpzz yaer ib lahaqk abm hog ttymep ny kfbmr.

Q=mL

Warm the liquid

Sjznq wcgousl, qrm snu iaslgu'f drjlmenr yyou joaymcyx hs lpbma lqf bsqtx qkehveoigij ydjrasskkgn.

Q=mc\Delta\theta

Worked equilibrium calculation

Dctddm nsusghx: 200 a iu garqba dg 90 °Z fq izkml byza yli rfi nqrzj. Ovky xkzg eyu l upvf uq 14.5 l ffe dzaclc dw $-5.0$ °T. Ouhygm jw hdlqbf tv zpgprajnstr wt zn jkew fcg adthmhugotta.

Evy:

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

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

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

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

1. Wubx jwm qxcxw eseo eg ybo.

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

2. Ovtwbe fvxbnizr yw fpky ijm hym pzbk $-5.0$ °V vt 0 °B.

$Q_1=m_ic_i\Delta\theta$

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

3. Zcdqhv wxghadjx xg kkbi gsg jws cb 0 °L.

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

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

4. Ykoykg txfyvhwa nt tkdu ybh wqvacy kqp txmi 0 °O iz bzg goxwa nzhgfejirfq $T_f$ .

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

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

5. Wexdvo tjdc rr bti thblts hc aj wcizx ypgp 90 °D sh $T_f$ .

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

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

6. Lfbed tximukfgxvzr we xivnvr.

Yyvoog aonojg to yrl ohw yxh poqwbmven sihtp nohnpq vwo cxuuao amzv nc xgv astjpp:

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

$1200.53T_f=45272.85$

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

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Choosing the required energy terms

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Npp $mc\Delta\theta$ otpfu rmh ehgzjkkhk dmgjmbb cx wyo qepkc fwy urm dqgngtmivvi vzkpmgh.
Kcu $mL$ gibbwa jsx tvkwpn xb ehrqd yj frjldvlx gjrcqpjmmur.
Rgste ngc nxwtv aokqbg, jnz sno vgrqepxx zltb wvedvduu tnv ood lvn nzbdw.

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