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### Homework Problem 8.21

Posted: Thu Jan 11, 2018 2:28 pm
Could someone please explain to me how to do homework problem 8.21, "A piece of copper of mass 20.0 g at 100.0 degrees Celsius is placed in a vessel of negligible heat capacity but containing 50.7 g of water at 22.0 degrees Celsius. Calculate the final temperature of the water. Assume that no energy is lost to the surroundings."
Thank you.

### Re: Homework Problem 8.21

Posted: Thu Jan 11, 2018 2:58 pm
All heat lost by the copper is going to be gained by the water. Therefore, $q_{metal}=-q_{water}$. $q=mC\Delta T$, where m=mass, C=specific heat capacity, and $\Delta T$ = the change in temperature.The specifc heat capacity of copper is $0.38 J.^{\circ}C^{-1}.g^{-1}$ and the specific heat capacity of water is$4.18 J.^{\circ}C^{-1}.g^{-1}$. So, you would solve $(20.0g)(T_{final}-100.0^{\circ}C)(0.38 J.^{\circ}C^{-1}.g^{-1})=-(50.7g)(T_{final}-22.0^{\circ}C)(4.18 J.^{\circ}C^{-1}.g^{-1})$.

### Re: Homework Problem 8.21

Posted: Thu Jan 11, 2018 3:08 pm
I did the same thing as Britney!! But some thing that might help while doing these problems is that specific heat values be found on the table on page 269, and you know that the final temperature will be somewhere in between the final and initial values, since one gains heat while the other loses heat.

### Re: Homework Problem 8.21

Posted: Thu Jan 11, 2018 9:26 pm
Just making sure, but is the final temperature of the water also the final temperature of the copper mass?

### Re: Homework Problem 8.21

Posted: Fri Jan 12, 2018 12:01 pm
Yes, the final temperature of the water is the same as the final temperature of the copper! It's because they are in the same container.