A sample of steam with a mass of 0.552 g and at a temperature of ...

Question

A sample of steam with a mass of 0.552 g and at a temperature of 
100 °C condenses into an insulated container holding 4.25 g of 
water at 5.0 °C. Assuming that no heat is lost to the surroundings, 
what is the final temperature of the mixture?

Accepted Answer

Hello everyone. So in this video we're dealing a lot with our Q. Equals M. C. Delta T. Equation. So the M. C. Delta T. Equation is also known as the crowd equation or the specific heat formula. So we have this Q. We have our cue equals M. C. Delta T. And this queue right here that stands for heat energy. Alright so before we begin this problem we're going to lay out some very key information. And so these values can be found either in your textbook or even given to you by a professor. So first I want to write that our delta H. Fusion. So I just put fus of water. So H two L. That's going to equal to six point oh one kg joules per mole. Then we have the specific heat of water. So H. 20. Is going to equal to 4.18 jewels programs, times degrees Celsius and then my molar mass of water. So H. 20. Is equal to 18. two g per mole. Alright, so first step is to calculate for my cue of ice. So maybe I'll put off to the side here. Alright, so first up, cute advice again, we're utilizing R. M. C. Delta T. Equation. We're specifically talking about just our ice. So the mass of that then is going to be one g. We're going to actually want to go ahead and convert our grams into moles using our molar mass and of course ice is just water. H. 20. So we'll have one more on top And 80.02 g on the bottom. Then we're gonna want to convert our most to killer jewels. So the unit of energy, we use that by the delta age of fusion of water. So we just have 6.01 kill jules per more. Now Lassie converting our killer jewels into jewels. So we'll have 10 to the third of jewels Her one killer jewel. And as you can see then the grams will cancel more will cancel and kill the jewels cancel leaving us with just our jewels and that's what we want. So put everything into my calculator, I'll get a total of 333. jewels. Alright not calculating for my cue of water, that's actually just equal to negative Q. Of ice. Again, Q of water is equal to negative 333. jewels. Alright. Next part. Now that we have this queue of H 20. We want to know each and every component specifically. We're interested in our delta T. So just kind of moving along to the side. Then I put the third step here. So again we'll put Q. F. H 20. Goes M. C. Delta T. So this time we have this QH. right over here let's go ahead and plug that in. We have negative 333. jewels. And in the mass we have that's going to be five g The specific heat capacity. So the C. We also have that's 4.18 jules per grams times degree Celsius. And we don't know our delta T. So solving just for our delta T. Just as if we were solving for the algebraic X. Then our delta T. Is equal to negative 15.9578°C. And of course now to solve for our temperature we know that delta T. Is equal to R. T. Final. So T. F. F minus T. F. I. Which is T. Initial. So we'll do 18 minus R. 15.9578. Of course this is all going to be in degrees Celsius. Putting putting this equation into my calculator I'll get that temperature is two point oh 4 to 2 degrees Celsius. Now that we have this crucial part of our problem has pace scroll down. Alright so last part then is calculating for my T. F. F. So basically rewriting this but in the opposite order you have cute of ice going to negative Q. Of H. 20. Again I'm just scrubbing a little bit more. Alright so kind of expanding out our Q. And what that means perhaps is R. M. C. T. Of F minus T. Initial. So ice equals to the negative M. C. And again T. F. F minus T. Of H. 20. Then putting actual values then your gets one g. And again of course the left side is talking about ice. Right side is talking about water Capacity 4.18 jules, Her grams times degree Celsius. Then we have our TFF -0°C Equals negative and the mass is going to be five g specific heat capacity is going to be seen as the ice. We have 4.18 joules programs times degree Celsius and of course, T of f -2.04-2°C. So we can see this is where this part came in. All right now we're solving for R T. F. F. That's kind of like solving for R. X. If you put this into our calculation, solve for essentially your ex here We'll get that T of F is equal to 1.7°C. And this is going to be our final answer for this problem. Thank you all so much for watching.

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