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Gabbruce
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)
What I know: Copper:155 g Water:250.0 g t: 150ºC-19.8ºC=130.2ºC How do I set up?
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?
Show me a little more please.
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?
Water is gaining heat and copper is losing heat
Sounds good. You know this because copper is hotter than water. Have you seen the "Heat Gained = - Heat Lost" equation before?
Have you heard of calorimetry?
Yes I have. Would the formula I have to use be heat gained by water= (mass)(specific heat)( ∆t)?
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?
Okay sounds good. I understand so far.
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?
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.
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.
I think Ti would be 130.2ºC
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.)
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)
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?
Okay. The problem is asking for the final temperature of the mixture.
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?
It's okay. Yes makes sense.
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?
Would i put my answers to the correct significant figure?
(1047x- 2.07x10^4)=59.675x-8.95x10^3 I do not think I wrote it to the correct significant figure.
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?
Can you rearrange now and solve for x?
Okay I see. 1106.175x=29671.95
x=26.82392027 x=2.682?
Before sig figs, I agree with first one x = 26.82392027
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.
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?
4.186... sorry dyslexia kicks in
But i still get the same answer as you did despite that minor difference. I did a google search to verify that number.
So what do you think the final answer will be with correct sf?
That would be the best answer ever...
Does this stuff make more sense to you now?
Yes much more sense! I was mainly having trouble creating the set up but now I understand. Thanks
Not a problem at all. Glad I could be of help.
Just one last question. Would it be 26.8J or no J?
The correct units would be degrees Celsius. This is because we are talking about temperature. Heat (Q) has units of J (joules)
Heat is energy, temperature is a relative measure of that energy so they are different things.
Okay makes sense thanks again!