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LanaderthneCertnidmeachonInitial rute:4L 00I (8q} T() ~ Ol(UTriallWlmMMnWlLsM4Q.o]E0.0832L0.450QUCJSI:JlWMsent7aeDeletntnNeaAmnrr BurJCCI |Delernunculuireidtion and...

Question

LanaderthneCertnidmeachonInitial rute:4L 00I (8q} T() ~ Ol(UTriallWlmMMnWlLsM4Q.o]E0.0832L0.450QUCJSI:JlWMsent7aeDeletntnNeaAmnrr BurJCCI |Delernunculuireidtion and Lhe correct unitsIperlloM-I

Lanaderthne Certnid meachon Initial rute: 4L 00I (8q} T() ~ Ol(U Triall Wlm MMn WlLs M4 Q.o]E 0.0832L 0.450 QUCJSI: JlW Mse nt7 ae Deletntn Nea Amnrr Bur JCCI | Delernun cului reidtion and Lhe correct units Iperllo M-I



Answers

$$ \begin{aligned} \Delta S=\frac{q_{m}}{T_{1}}+c \ln \frac{T_{2}}{T}+\frac{q_{v}}{T_{2}} & \\ &=\frac{333}{273}+4 \cdot 18 \ln \frac{373}{283} \sim 8.56 \mathrm{~J} /{ }^{\circ} \mathrm{K} \end{aligned} $$

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.

Hello students in this question we have mass of water. M equals to one kg. And which is heated from the temperature T. One equals to 10 degrees integrate up to 100 degrees integrate that is temperature T to. Okay at which it evaporate completely. Okay evaporation is also taking place. Okay now we have to find the entropy increments of the system. So here we have two processes. One is the temperature change and one is the phase change. Okay so changing entropy will be given by Yeah during the phase change the entropy is EMC Ellen of temperature T. To develop it temperature T even Ok and during the face change the entropy is given by mm latent heat of the uh the tradition diver be temperature T to Okay no mm which is one kg and see which is 4.18. Multiply by 10 to the power three jewel per program Calvin and natural log of T two which is 100 degrees integrated that is 3 73 Calvin and 10 degrees integrate which is 2 83 Calvin. Okay So 3 73 divided by 283 Calvin plus M. Which is one kg. And latent heat of Where projection is double to 50 player they tend to the power three kill a jewel per kg, develop a temperature T two which is 3 73 kelvin. So after solving the increment indian therapy is obtained as 7.2 more player beaten to the power three jewel per kelvin or it can be written as 7.2 kg per kelvin. Okay so this become answered for the problem. Okay, thank you.

So here we have the integral of 12 t to the eighth Dynasty all over t to the power of three halves. DT. So what we're going to do here is we're going to divide uh uh T to the three halves into the 12 t to the eighth and the minus t using the rules of exponents of the laws of exponents. So when you divide uh, to two bases to like bases, you subtract the exponents, right? So that's what we're gonna do here. So first, we're going to split they in a group as the integral of 12 t to the eighth, divided by t to the three halves DT minus the integral of tea, divided by t to the three house DT, and so 12 comes out of the integral. And then here we can just subtract the exponents using a lot that's one of the laws of exponents, right? Division of like bases means you can subtract the exponents. So becomes teacher. The power of eight minus three halves, Bt minus. Now, here there's any visible exponents of one. So you have the integral of tea to the one minus three halves. DT so that gives us 12 in a growth now. Eight minus three halves. Let's see. Eight is 16 halves minus three halves will be 13. Have so t to the 13. Have the team minus. And then here a t two. Let's see. One is to have so two minus three negative. 1/2 Bt. Okay, so let's go over here. So now we have, what, 12 times? T. So here, we're gonna apply the power room. Right? So applying the power rule gives us t to the power of 13/2 plus one. Guy had the one the right by the same exponents minus same thing here that supply the power right there. So t to the negative 1/2 plus one divided by a negative 1/2 plus one plus the arbitrary constant of integration for this becomes 12 uh, 13/2 plus 2/2 is 15/2, divided by 15/2 minus negative, 1/2 plus two halves. This 1/2 so t to the 1/2 divided by 1/2 plus c and then we can clean things up a bit. So 12 divided by 15/2 is 12 uh, times. Uh, to over 15 T to the 15 halves minus. Uh, here Dividing something by 1/2 is the same thing. It's multiplying. That's something by choose. Becomes too a t to the 1/2 plus c. So what? We get 12 and 15 have a common factor of 3 12 Divided by three is 4 15 better by threes. 54 times two is eight to have 8/5. Uh, T to the 15 halves minus two. T two. The 1/2 t started about my hand writing here, um, policy and that steals.

In this problem we will cover the differential of two variable function. So I have written in Green the general equation for differential. So we know that in the context of this problem we will be finding the partial derivatives of the function G with respect to you envy. So I will start with the partial derivative with respect to you. And that's just taking the derivative folding be fixed. So we will get to you plus we and we want the partial derivative of respected me. And that's just taking the derivatives holding you fixed. And this is going to give us just you. So now that we have our partial derivatives, we can write out our differential and that's going to be D. S equals two U plus V. Do you wait my mistake? This is not enough. It's a G plus you times DV. And there goes our differential. So now we want to estimate the change between the two given points. And to do that, we can expand or differential to look like this. So we have D G equals Let's see, two U plus B Times U -1 -1 plus you Times V -2. And for our partial derivatives which you're here in here, we are going to use the point 12 So we're going to rewrite this as G is equal to let's see that's going before u minus one plus one times being an a student. So that's just going to be b minus two. So now that we have that we are going to now plug in our second point, which is 1.22.1 so D G Is equal four times 1.2 -1 Plus 2.1 minus two. That equals four times 0.2 plus zero point one. And that is going to give us and the answer of zero 0.9 and this should be our estimated change in the value of function G as you move from the first point to the second.


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