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Ch 20: The Second Law of Thermodynamics

Chapter 20, Problem 20

A sophomore with nothing better to do adds heat to 0.350 kg of ice at 0.0°C until it is all melted. (a) What is the change in entropy of the water?

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welcome back everybody. We are taking a look at a cube of butter and we are told that this butter is going to melt. We're also told the mass of the butter cube is 6.9 g and that it has a melting point of 35 degrees Celsius. We're also told that it has a latent heat of fusion Of 60 joules per gram. And we are asked with finding what the change in entropy is once the butter melts. But we have a formula for this. The change in entropy is equal to the heat that leaves over. Let's see here the temperature of the melting point of the butter. But we don't have cue yet. So Q. We have another formula four is just mass times the latent heat of fusion. So let's go ahead and find that real quick. We have that are mass is 6.9 g times 60 jewels per gram. The units of graham is going to cancel out, leaving us with 414 jewels of heat. Great. So now let's go ahead and plug in our values into our delta S. Formula. We have 414 jewels divided by degrees Celsius. But we need this in Calvin. So I'm gonna add 273.15 to convert it to Calvin. When you plug this into your calculator, we get that a change in entropy is 1.34 jewels per kelvin, which corresponds to our final answer of a thank you all so much for watching. Hope this video helped. We will see you all in the next one.