Welcome to PHY 121 Blog Help. Here's how it works. For each homework question and lab report we will make a post, this will probably contain a few tips on what the problems are about and how to solve them. If you are stuck on something then instead of emailing us directly you should post a comment in reply to the relevant post. We will try to guide you through tough points and help you understand the problems and the concepts behind them.
Friday, November 26, 2010
Ch 13_2 #3
Problem similar to the problem Ch 13_2 #2
14 comments:
Anonymous
said...
I've tried using Q=m*c*deltaT I set Q as 220*4.186*81 (same thing i did for the previous problem, which worked), and got 85,821 J. I divided this value by the specific heat of ice (2090) and then by 42 degrees (desired change in temp).
This didn't work. Can someone help me understand why?
I think it would help if we did some lecture on this material before the problems are due, rather than having the homework due without having any knowledge of how to solve the problems.
Someone has to really watch one's language if in the future they want to get help with homework problems. If one can not ask an intelligent question may be should not be really be taking this course.
For the rest of you who follow this discussion this problem is similar to problem Ch 13_2 #2 as it uses the same physical principles. That is heat gained = heat lost and the final temperature for all object is the same. Using those 2 principles write down 2 equations for heat one for coffee and one for ice and solve it for the unknown quantity.
i did Q=mcdelta t and got 230*4.186*88 and then divided by 290 and 32 and got 1.2668. Then I added this with anouther 230*3.3*10^5 with the help of the last post and got 75900001.27 and its wrong again and i dont understand where i made the mistake
Since everybody set has different numerical answers it hard to see what goes wrong. It would be much clearer if people described in words what they looking for in intermediate steps and wrote formulas in generic variables rather that numbers. In post #10 what is 290 why division happening at all? Why the heat from the melted ice added? Should that heat be subtracted since you spend it melting ice?
14 comments:
I've tried using Q=m*c*deltaT
I set Q as 220*4.186*81 (same thing i did for the previous problem, which worked), and got 85,821 J.
I divided this value by the specific heat of ice (2090) and then by 42 degrees (desired change in temp).
This didn't work. Can someone help me understand why?
I think it would help if we did some lecture on this material before the problems are due, rather than having the homework due without having any knowledge of how to solve the problems.
ahahhhhaaaaaaaaa comment #2 you made my night
oh mannn....
Someone has to really watch one's language if in the future they want to get help with homework problems. If one can not ask an intelligent question may be should not be really be taking this course.
For the rest of you who follow this discussion this problem is similar to
problem Ch 13_2 #2 as it uses the same physical principles. That is heat gained = heat lost and
the final temperature for all object is the same. Using those 2 principles
write down 2 equations for heat one for coffee and one for ice and solve it for the unknown quantity.
For comment #1. You seem to be using Delta T = 81 C. Why? What is the final temperature of coffee?
What the professor failed to mention is that this problem also involves a phase change. Thus, your equation for ice should read:
Q = m*Lf + m*c*DeltaT
Find Q with the water equation, factor out the m from the ice equation, and solve for m. Lf is a constant; when using kg its 3.33 *10^5
i did Q=mcdelta t and got 230*4.186*88 and then divided by 290 and 32 and got 1.2668. Then I added this with anouther 230*3.3*10^5 with the help of the last post and got 75900001.27 and its wrong again and i dont understand where i made the mistake
If you follow the equation on 18'and isolate the mass for ice, you will arrive at the answer.
Since everybody set has different numerical answers it hard to see what goes wrong. It would be much clearer if people described in words what they looking for in intermediate steps and wrote formulas in generic variables rather that numbers. In post #10 what is 290 why division happening at all? Why the heat from the melted ice added? Should that heat be subtracted since you spend it melting ice?
How many significant figures do I have to use? It is asking for more significant figures than are in the given values.
It does not specify so I would enter 3 or 4.
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