June 25, 2024, 10:41:40 PM
Forum Rules: Read This Before Posting


Topic: Thermochemistry problem  (Read 5515 times)

0 Members and 1 Guest are viewing this topic.

Offline Johan

  • Regular Member
  • ***
  • Posts: 13
  • Mole Snacks: +0/-0
Thermochemistry problem
« on: November 13, 2007, 10:50:57 PM »
325.0g piece of gold at 427 C is dropped into 200.0ml of water at 22.0 C. Calculate the final temperature of the mixture. Specific Heat of the Gold = .0131J/gC

So confused.. I know how to get grams of water from the equation.. but i dunno how to relate the rest of the information.

I tried setting qH2O= qgold but it just wont work.. i know the q from the gold is going to be negative.. but yea.. I can't calculate q individually because i can't find the change in temperature so what do i do?

Offline LQ43

  • Chemist
  • Full Member
  • *
  • Posts: 250
  • Mole Snacks: +32/-9
  • Gender: Female
Re: Thermochemistry problem
« Reply #1 on: November 13, 2007, 11:33:08 PM »
qH2O= qgold

this is a good idea

q = Cp x m x (Tf-Ti)

the Tf is what you are looking for and is the same in both q 's

1 equn 1 unknown

you know all other variables, just remember to use the correct Cp for the right m

Offline Johan

  • Regular Member
  • ***
  • Posts: 13
  • Mole Snacks: +0/-0
Re: Thermochemistry problem
« Reply #2 on: November 14, 2007, 03:17:02 AM »
Yea i'm totally stumped. I have the same equation.. I can't find q without change of temperature, nor can i find specific heat without it.. unless i'm completely missing something.

Do i ignore q and set the rest of it to zero just solving for Tf?

That doesn't seem right at all.. i need a value for q and Cp, i'm so lost :( please help. Cp is Heat Capacity under constant pressure isn't it? If it is.. then i really have no idea what to do.

Offline AlIXe

  • New Member
  • **
  • Posts: 8
  • Mole Snacks: +0/-0
Re: Thermochemistry problem
« Reply #3 on: November 14, 2007, 03:51:34 AM »
Both substances will have the same final temperature. They will have different values of q and evidently different starting temperatures

Try:

q(Gold) = .0131J/g°C * 325 * (T(final) - 427)

q(water) = 4.186 J/g°C * 200g * (T(final) - 22)

Solve both values of q and T final by simultaneous equation
Hope this helps

Offline Padfoot

  • Full Member
  • ****
  • Posts: 221
  • Mole Snacks: +23/-2
Re: Thermochemistry problem
« Reply #4 on: November 14, 2007, 04:14:40 AM »
Both substances will have the same final temperature. They will have different values of q and evidently different starting temperatures

Try:

q(Gold) = .0131J/g°C * 325 * (T(final) - 427)

q(water) = 4.186 J/g°C * 200g * (T(final) - 22)

Solve both values of q and T final by simultaneous equation
Hope this helps

On the right track, but Heat absorbed by water = heat lost by gold
so: q(gold)=q(water)

The 2 eqns you have above can combine - subbing in q(water) for q(gold) for example to get one eqn - with one unknown (Tfinal).

To Johan: I think this was the point LQ43 was initially making


Offline AlIXe

  • New Member
  • **
  • Posts: 8
  • Mole Snacks: +0/-0
Re: Thermochemistry problem
« Reply #5 on: November 14, 2007, 08:42:59 AM »
yes, sorry i wasn't to clear, when i said different values of q i meant q(water)= - q (gold)

hope this helps

Sponsored Links