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Chemistry: 2 samples of the same liquid but different temp - final temp?

by meganw
Tags: chemistry, final, liquid, samples, temp
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meganw
#1
Nov1-08, 02:44 PM
P: 96
1. The problem statement, all variables and given/known data

75 mL of a pure liquid at 245 K is mixed with 100 mL of the same pure liquid 365 K. What is the final temperature?

The answer is 314 K but I'm not sure why!

Thank you! =)

2. Relevant equations

It says this problem has to do with Calorimetry but I'm not sure why! Calorimentry equations involve q=mc(d)t...but how would you incorporate both liquids?

3. The attempt at a solution

q=mc(d)t? I'm so lost, sorry!
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hage567
#2
Nov1-08, 05:32 PM
HW Helper
P: 1,539
Yes you need that equation. Energy is conserved. You need to remember that the heat lost by the water at 365 K is equal to the heat gained by the volume of water at 345 K.
meganw
#3
Nov1-08, 09:50 PM
P: 96
I'm sorry I'm afraid I still don't understand...how can I use q=mcat for this problem?

hage567
#4
Nov1-08, 10:18 PM
HW Helper
P: 1,539
Chemistry: 2 samples of the same liquid but different temp - final temp?

That equation tells you the amount of thermal energy that is transferred between a mass m and its surroundings to undergo a temperature change ∆T.

So, you have Q1 = m1c∆T for the first volume of water, and Q2 = m2c∆T for the second volume of water.
When they are mixed, the heat lost by the water at the higher temperature will equal the heat gained by the water at the lower temperature. They will come to equilibrium at the final temperature, Tf.
meganw
#5
Nov1-08, 11:13 PM
P: 96
But how can you set it up?

I tried mcat=mcat but you get 75=100 which is obviously not true..? Can you pleas explain how to solve the q=mcat problem? I'm really sorry I just don't get it...
hage567
#6
Nov1-08, 11:47 PM
HW Helper
P: 1,539
I tried mcat=mcat but you get 75=100 which is obviously not true
You can't eliminate the ∆Ts, they aren't the same. Do you know what ∆T means? It is the difference between the final and initial temperatures of the liquid, so ∆T = Tf - Ti. Knowing this, you can complete your equations Q1 and Q2. What's the expression for ∆T for the water at 245 K? What's ∆T for the water at 365 K? You know the initial temperature for each, and you want to find Tf which will be the equilibrium temperature after the two mix together. And as I said above, energy is conserved, so Q1 + Q2 = 0. You need to solve this equation for Tf.

See if that helps. I can't say much more without just giving you the solution, so try to think through what I've said and see how far you can get.
meganw
#7
Nov2-08, 07:57 AM
P: 96
Okay, thank you!!

I ended up getting 312.143 K which is a little off from 314. Did you guys get 314 or 312.1 like me?

75(Final-245) + 100(Final-365)=0 ("c" cancels)
75x -183675 + 100x - 36500 = 0 (let "x" = temp final)
175x-54625=0
x=312.143
hage567
#8
Nov2-08, 09:12 AM
HW Helper
P: 1,539
75x -183675 + 100x - 36500 = 0 (let "x" = temp final)
175x-54625=0
Check your math carefully. You have an extra number in the 183675 but I think it might be a typo, so I'm not sure where you went wrong. I get 314 K after rounding.
meganw
#9
Nov2-08, 09:31 AM
P: 96
Opps, yes! A typ-o!

And I did get 313.6 when I tried it again. Thank you so much for your continued help and patience!! =) I really appreciate it because I know you didn't have to do it but you were willing to stick with me until I got it, which was really nice of you. Thanks again.
hage567
#10
Nov2-08, 10:50 AM
HW Helper
P: 1,539
You're welcome.


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