anonymous
  • anonymous
a 155 gram sample of copper was heated to 150.0ºC then placed into 250 gram of water at 19.8ºC. (The specific heat of copper is 0.385J/gºC.)Calculate the final temperature of the mixture. (assume no heat loss to the surroundings)
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
What I know: Copper:155 g Water:250.0 g t: 150ºC-19.8ºC=130.2ºC How do I set up?
anonymous
  • anonymous
You want to set it up such that Heat Gained = -Heat Lost. Have you heard of this or do you want me to show you a little more?
anonymous
  • anonymous
Show me a little more please.

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anonymous
  • anonymous
Sure. I'll ask you a series of questions so you hopefully understand the process. Which is gaining heat and which is losing heat? Copper or Water & how do you know?
anonymous
  • anonymous
Water is gaining heat and copper is losing heat
anonymous
  • anonymous
Sounds good. You know this because copper is hotter than water. Have you seen the "Heat Gained = - Heat Lost" equation before?
anonymous
  • anonymous
no
anonymous
  • anonymous
Have you heard of calorimetry?
anonymous
  • anonymous
Yes I have. Would the formula I have to use be heat gained by water= (mass)(specific heat)( ∆t)?
anonymous
  • anonymous
Yes but you use it twice. Let me explain. You have 2 things. One is clearly hot and the other is cold. in an isolated system (meaning those 2 things are the only things that interact with one another) the thermal energy is going to be exchanged only between those two objects and nothing else. This is an ideal case. You told me that the water will gain heat and the copper will lose heat. This being the case, the total thermal energy between the two objects does not change (according to the conservation of energy - energy is never created or destroyed, only transferred from one form to another). So if we set this up in the form of an equation we can see: \[Q _{Lost} + Q _{Gained} = 0\] \[Q _{Gained} = -Q _{Lost}\] But because the copper is losing and the water is gaining: \[Q _{Water} = -Q _{Copper}\] I will show you how to use the mC Delta t stuff next. Do you have any questions so far?
anonymous
  • anonymous
Okay sounds good. I understand so far.
anonymous
  • anonymous
Ok. Now lets talk about your mass, specific heat and temp change by defining the variables. Q stands for heat and has units of Joules (J) m stands for mass and has units of (g) C stands for specific heat and has units of (J/(g deg C)) Tf is the final temperature of a substance (deg C) Ti is the initial temperature of a substance (deg C) The equation for a given heat calculation is: \[Q = mC \Delta T\] Which is also written as \[Q = mC(T _{f}-T _{i})\] Still ok?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Ok. What we are going to do is make a list of what we know and what we are trying to find. I want you to tell me the m, C, Ti, and Tf for both copper and water. Tell me what we are trying to find out of those variables as well. If you dont know something or have a number that you think you should, let me know and we'll talk about it. Then, once this is taken care of, we will set up our master equation that looks like an algebra problem to solve and its pretty straight forward from there.
anonymous
  • anonymous
m (copper): 155g m (water): 250.0g C: 0.385 J/gºC Ti (copper): 150.0ºC (not sure if this is consider ti) Ti (water): 19.8ºC (not sure also) Tf: We are trying to solve for this. The final temperature of the mixture.
anonymous
  • anonymous
I think Ti would be 130.2ºC
anonymous
  • anonymous
I like what I see above for your initial conditions but I do have a few questions for you. What is C for water? What do you mean Ti would be 130.2? (How do you do degree symbol? I'm curious.)
anonymous
  • anonymous
I do not know what C is for water. For Ti, what I did was 150.0ºC-19,8ºC but i guess that is wrong. So i am assuming what I wrote above is correct: Ti (copper): 150.0ºC (not sure if this is consider ti) Ti (water): 19.8ºC (not sure also) (For the degree symbol you hold alt and click on zero. º)
anonymous
  • anonymous
Thanks for the tip.
anonymous
  • anonymous
You welcome.
anonymous
  • anonymous
The C for water is not provided. This is a standardized value that can be found in the internet. Typically C for water is 4.816 J/(gC). We will use this number. Do not subtract any numbers or do anything at this point for Ti. We are just seeing what we have. What is the problem asking us to find?
anonymous
  • anonymous
Okay. The problem is asking for the final temperature of the mixture.
anonymous
  • anonymous
Yes you are correct. (sorry pc started to get slow on me...) We will call Tf = x. We do not know this for either material. This is the point where we set up our algebra problem as shown below. \[Q _{Water} = -Q _{Copper}\] \[m _{w}C _{w}(T _{f}-T _{i})_{w} = -m _{cu}C _{cu}(T _{f}-T _{i})_{cu}\] That make sense so far?
anonymous
  • anonymous
It's okay. Yes makes sense.
anonymous
  • anonymous
Ok. Now we plug in numbers. \[(250)(4.186)(x-19.8)=-(155)(0.385)(x-150)\] Can you reduce this down on both sides and tell me what you get before solving for x?
anonymous
  • anonymous
Would i put my answers to the correct significant figure?
anonymous
  • anonymous
Yes please.
anonymous
  • anonymous
(1047x- 2.07x10^4)=59.675x-8.95x10^3 I do not think I wrote it to the correct significant figure.
anonymous
  • anonymous
Oh I'm sorry, I thought you were giving me an answer. Typically, I never go to correct sig figs until the very end. This is very close to what I have. Let me show you what I have. \[1046.5x -20720.7 = -59.675x + 8951.25\] Sound good so far?
anonymous
  • anonymous
Can you rearrange now and solve for x?
anonymous
  • anonymous
Okay I see. 1106.175x=29671.95
anonymous
  • anonymous
keep going.
anonymous
  • anonymous
x=26.82392027 x=2.682?
anonymous
  • anonymous
Before sig figs, I agree with first one x = 26.82392027
anonymous
  • anonymous
I see 3 sf in all of the variables. I do not consider constants such as specific heat of water in my sf calculation because it is known to be exactly what the number is unless otherwise indicated. Mass for both quantities have 3 sf (assuming it is 250g exactly). Same goes for Ti, 3sf as long as it is 150 exactly.
anonymous
  • anonymous
Wait I have a question. You mentioned water is 4.816 J/(gºC) but in the equation you wrote water is 4.186. Which is the correct one?
anonymous
  • anonymous
4.186... sorry dyslexia kicks in
anonymous
  • anonymous
But i still get the same answer as you did despite that minor difference. I did a google search to verify that number.
anonymous
  • anonymous
So what do you think the final answer will be with correct sf?
anonymous
  • anonymous
2.68?
anonymous
  • anonymous
You mean 26.8?
anonymous
  • anonymous
Oh yes, 26.8
anonymous
  • anonymous
That would be the best answer ever...
anonymous
  • anonymous
Does this stuff make more sense to you now?
anonymous
  • anonymous
Yes much more sense! I was mainly having trouble creating the set up but now I understand. Thanks
anonymous
  • anonymous
Not a problem at all. Glad I could be of help.
anonymous
  • anonymous
Just one last question. Would it be 26.8J or no J?
anonymous
  • anonymous
The correct units would be degrees Celsius. This is because we are talking about temperature. Heat (Q) has units of J (joules)
anonymous
  • anonymous
Heat is energy, temperature is a relative measure of that energy so they are different things.
anonymous
  • anonymous
Okay makes sense thanks again!
anonymous
  • anonymous
No problem. Later.
anonymous
  • anonymous
Later.

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