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

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SUMMARY

The final temperature of a mixture of 75 mL of liquid at 245 K and 100 mL at 365 K is calculated to be 314 K using the principles of calorimetry. The heat lost by the warmer liquid equals the heat gained by the cooler liquid, expressed through the equation Q1 + Q2 = 0. The specific heat equation q = mc∆T is applied, where the mass (m) and specific heat (c) cancel out, allowing for the determination of the final temperature (Tf) through the temperature differences. The calculations confirm that the final equilibrium temperature is approximately 314 K.

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meganw
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Homework Statement



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! =)

Homework 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?

The Attempt at a Solution



q=mc(d)t? I'm so lost, sorry!
 
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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.
 
I'm sorry I'm afraid I still don't understand...how can I use q=mcat for this problem?
 
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.
 
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...
 
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.
 
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
 
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.
 
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. :smile:
 
  • #10
You're welcome. :smile:
 

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