4A.9

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205150314
Posts: 106
Joined: Wed Feb 20, 2019 12:16 am

4A.9

Postby 205150314 » Tue Jan 28, 2020 9:49 pm

Im getting a negative answer for heat released = heat absorbed, does anyone know why?
im doing 20*.39*(100-T)= 50.7*4.18(T-22)

Justin Vayakone 1C
Posts: 110
Joined: Sat Sep 07, 2019 12:19 am

Re: 4A.9

Postby Justin Vayakone 1C » Tue Jan 28, 2020 10:15 pm

Heat released = - heat absorbed
Your work here looks right. I'm assuming you used the negative to change (T-100) into (100-T) right?

Hui Qiao Wu 1I
Posts: 56
Joined: Fri Aug 30, 2019 12:16 am

Re: 4A.9

Postby Hui Qiao Wu 1I » Wed Jan 29, 2020 3:26 pm

Your change in temperature for copper should be final temperature-100, not the other way around.

Katie Kyan 2K
Posts: 106
Joined: Fri Aug 09, 2019 12:16 am
Been upvoted: 1 time

Re: 4A.9

Postby Katie Kyan 2K » Wed Jan 29, 2020 3:49 pm

You calculated the change in temperature wrong. It should be final temperature - 100.

Anisha Chandra 1K
Posts: 118
Joined: Thu Jul 11, 2019 12:17 am

Re: 4A.9

Postby Anisha Chandra 1K » Thu Jan 30, 2020 11:12 am

Yeah, you can do (Tfinal - 100) and then add a negative sign to the other side with water to show that the heat lost by copper is that gained by water.


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