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anonymous
 4 years ago
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)
anonymous
 4 years ago
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)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What I know: Copper:155 g Water:250.0 g t: 150ºC19.8ºC=130.2ºC How do I set up?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You 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
 4 years ago
Best ResponseYou've already chosen the best response.0Show me a little more please.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sure. 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
 4 years ago
Best ResponseYou've already chosen the best response.0Water is gaining heat and copper is losing heat

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sounds good. You know this because copper is hotter than water. Have you seen the "Heat Gained =  Heat Lost" equation before?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Have you heard of calorimetry?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes I have. Would the formula I have to use be heat gained by water= (mass)(specific heat)( ∆t)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes 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
 4 years ago
Best ResponseYou've already chosen the best response.0Okay sounds good. I understand so far.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok. 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
 4 years ago
Best ResponseYou've already chosen the best response.0Ok. 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
 4 years ago
Best ResponseYou've already chosen the best response.0m (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
 4 years ago
Best ResponseYou've already chosen the best response.0I think Ti would be 130.2ºC

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0I do not know what C is for water. For Ti, what I did was 150.0ºC19,8ºC 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
 4 years ago
Best ResponseYou've already chosen the best response.0The 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
 4 years ago
Best ResponseYou've already chosen the best response.0Okay. The problem is asking for the final temperature of the mixture.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes 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
 4 years ago
Best ResponseYou've already chosen the best response.0It's okay. Yes makes sense.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok. Now we plug in numbers. \[(250)(4.186)(x19.8)=(155)(0.385)(x150)\] Can you reduce this down on both sides and tell me what you get before solving for x?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Would i put my answers to the correct significant figure?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(1047x 2.07x10^4)=59.675x8.95x10^3 I do not think I wrote it to the correct significant figure.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh 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
 4 years ago
Best ResponseYou've already chosen the best response.0Can you rearrange now and solve for x?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay I see. 1106.175x=29671.95

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x=26.82392027 x=2.682?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Before sig figs, I agree with first one x = 26.82392027

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0Wait 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
 4 years ago
Best ResponseYou've already chosen the best response.04.186... sorry dyslexia kicks in

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But i still get the same answer as you did despite that minor difference. I did a google search to verify that number.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So what do you think the final answer will be with correct sf?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That would be the best answer ever...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Does this stuff make more sense to you now?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes much more sense! I was mainly having trouble creating the set up but now I understand. Thanks

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Not a problem at all. Glad I could be of help.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Just one last question. Would it be 26.8J or no J?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The correct units would be degrees Celsius. This is because we are talking about temperature. Heat (Q) has units of J (joules)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Heat is energy, temperature is a relative measure of that energy so they are different things.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay makes sense thanks again!
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