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Question

Me anarnenteyeMdant ' ="Tal IN runtat4 & htananimaatFom Mxknt [3ld044704*n [4raten Gto Mune [email protected] [arean mamSALD D S0anitu maOeeanenEemn0--=!--e! --=1TENoTlnn2Fnaain GeenRann cn(eKI)UOa -I)CI)Uqnr Coaeen tanal (I+ (-I) = +-I+(-mWro mxFoaeaa(KV) + (K-W) + (-8J(I) + (-VJ(-W)= EI-I-It6--"-C FI+(-IK-1) 3(-(-4-1QD.Lidtt )-Contka 04830x tOro Fmmn nuLro men Taeten Gen15m"/Wnaha Juou @~Ua et nrseto ceduttrncuehadtaxuuumb Ia eortimnFnlon shaudbe %bled skte t [Meenontarnnnta enomo

Me anar nenteye Mdant ' = "Tal IN runtat4 & htananimaatFom Mxknt [3ld044704*n [4raten Gto Mune [email protected] [arean mamSALD D S0anitu ma Oeeanen Eemn 0--=!--e! --=1 TENo Tlnn2 Fnaain Geen Rann cn (eKI) UOa -I)CI) Uqnr Coaeen tanal (I+ (-I) = +-I+(-m Wro mx Foae aa (KV) + (K-W) + (-8J(I) + (-VJ(-W)= EI-I-It6--"-C FI+(-IK-1) 3(-(-4-1QD. Lidtt )- Contka 04830x tOro Fmmn nuLro men Taeten Gen 15m"/ Wnaha Juou @~Ua et nrs eto ceduttrn cuehadtaxuuumb Ia eortimn Fnlon shaudbe %bled skte t [ Meenontarnn nta enomov: (~J7 M ln e1 €eaTlate ea bonpgzansadoted Jannd04ecotr tat Men Dratiatat D4Sdn | Gndenne Or Krden O Enden J Baadiin7 Delicdant S4ana Otsucnzar 5eoa ] Aoidam ton&mmtaraa



Answers

$$ \begin{aligned} &\text { Here } Q_{1}^{\prime \prime}=C_{p} T_{0}(\tau-1), Q^{\prime} 1^{\prime \prime}=\tau R T_{0} \ln n \text { and }\\ &Q_{2}^{\prime \prime}=C^{p} T_{0}(\tau-1), Q_{2}^{\prime \prime \prime}=R T_{0} \ln n\\ &\text { in addition to we have }\\ &Q_{1}=Q_{1}^{\prime}+Q_{1}^{\prime \prime} \text { and }\\ &\begin{aligned} &Q_{2}^{\prime}=Q_{2}^{\prime \prime}+Q_{2}^{\prime \prime \prime} \\ &\text { So } \eta=1-\frac{Q_{2}^{\prime}}{Q_{1}}=1-\frac{C_{p}(\tau-1)+R \ln n}{C_{p}(\tau-1)+\tau R \ln n} \\ &=1-\frac{\tau-1+\left(1-\frac{1}{\gamma}\right) \ln n}{\tau-1+\left(1-\frac{1}{\gamma}\right) \tau \ln n} \\ &=1-\frac{\tau-1+\left(1-\frac{1}{\gamma}\right) \ln n}{\tau-1+\left(1-\frac{1}{\gamma}\right) \tau \ln n}=\frac{(\tau-1) \ln n}{\tau \ln n+\frac{\gamma(\tau-1)}{\gamma-1}} \end{aligned} \end{aligned} $$

Some of the 1st and 2nd largest energies and those of 3rd and 4th energies and I are okay, so and I. And the beauty. So first are you 1913 seconds And I have electronic confusion. 38 for us to It has also 5 38 six Astros. So in both are belong to a group. In a group. Are your nine senators will be please. So Lincoln has high. Are your major energy complex is formed by a night, correct up some unity. See

Hello students in this question we have molten lead of mass m equals to five g at temperature T two equals 23 27 degrees integrate And this is equal to 600 Calvin. OK and them which is the melting point temperature of the late which is poured into the calorie meat package. Which large amount of eyes having temperature T one equals to zero degrees integrate that is 37 to 73. So we have to find the entropy increments of the system that is led ice by the movement that thermal equilibrium is reached. Okay so we can right here that change in entropy delta. S will be given by heat released by the leader. So this will be given by Emma claire by Cuban. Do better be develop a. This temperature T two minus M. C. Ellen of temperature T. To develop it even this is during the face change and this is during the temperature change less I am of Q two Plus C. and Delta T. They will be temperature even. Okay so this can be rearranged to write that delta as it is equal to the delta as it is equals two molecular break. You too. This is also cute which is for the eyes. Okay so M. Q two and one by temperature T even minus one by temperature T. Two and plus Ember player B. C. And this is T to bite even minus one minus T. To buy even. Okay so this will be canceled out. So we can substitute all the values so delta S will be equals two after substituting values double 245 Plus 0.2564. And this is equals to the 0.48 jewel per Calvin. Okay, so this is the answer for the problem. Okay, Thank you.


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