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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)

  • one year ago
  • one year ago

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  1. Gabbruce
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    What I know: Copper:155 g Water:250.0 g t: 150ºC-19.8ºC=130.2ºC How do I set up?

    • one year ago
  2. ScienceGuy
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    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?

    • one year ago
  3. Gabbruce
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    Show me a little more please.

    • one year ago
  4. ScienceGuy
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    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?

    • one year ago
  5. Gabbruce
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    Water is gaining heat and copper is losing heat

    • one year ago
  6. ScienceGuy
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    Sounds good. You know this because copper is hotter than water. Have you seen the "Heat Gained = - Heat Lost" equation before?

    • one year ago
  7. Gabbruce
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    no

    • one year ago
  8. ScienceGuy
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    Have you heard of calorimetry?

    • one year ago
  9. Gabbruce
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    Yes I have. Would the formula I have to use be heat gained by water= (mass)(specific heat)( ∆t)?

    • one year ago
  10. ScienceGuy
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    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?

    • one year ago
  11. Gabbruce
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    Okay sounds good. I understand so far.

    • one year ago
  12. ScienceGuy
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    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?

    • one year ago
  13. Gabbruce
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    Yes

    • one year ago
  14. ScienceGuy
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    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.

    • one year ago
  15. Gabbruce
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    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.

    • one year ago
  16. Gabbruce
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    I think Ti would be 130.2ºC

    • one year ago
  17. ScienceGuy
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    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.)

    • one year ago
  18. Gabbruce
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    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)

    • one year ago
  19. ScienceGuy
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    Thanks for the tip.

    • one year ago
  20. Gabbruce
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    You welcome.

    • one year ago
  21. ScienceGuy
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    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?

    • one year ago
  22. Gabbruce
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    Okay. The problem is asking for the final temperature of the mixture.

    • one year ago
  23. ScienceGuy
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    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?

    • one year ago
  24. Gabbruce
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    It's okay. Yes makes sense.

    • one year ago
  25. ScienceGuy
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    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?

    • one year ago
  26. Gabbruce
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    Would i put my answers to the correct significant figure?

    • one year ago
  27. ScienceGuy
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    Yes please.

    • one year ago
  28. Gabbruce
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    (1047x- 2.07x10^4)=59.675x-8.95x10^3 I do not think I wrote it to the correct significant figure.

    • one year ago
  29. ScienceGuy
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    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?

    • one year ago
  30. ScienceGuy
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    Can you rearrange now and solve for x?

    • one year ago
  31. Gabbruce
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    Okay I see. 1106.175x=29671.95

    • one year ago
  32. ScienceGuy
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    keep going.

    • one year ago
  33. Gabbruce
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    x=26.82392027 x=2.682?

    • one year ago
  34. ScienceGuy
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    Before sig figs, I agree with first one x = 26.82392027

    • one year ago
  35. ScienceGuy
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    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.

    • one year ago
  36. Gabbruce
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    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?

    • one year ago
  37. ScienceGuy
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    4.186... sorry dyslexia kicks in

    • one year ago
  38. ScienceGuy
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    But i still get the same answer as you did despite that minor difference. I did a google search to verify that number.

    • one year ago
  39. ScienceGuy
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    So what do you think the final answer will be with correct sf?

    • one year ago
  40. Gabbruce
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    2.68?

    • one year ago
  41. ScienceGuy
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    You mean 26.8?

    • one year ago
  42. Gabbruce
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    Oh yes, 26.8

    • one year ago
  43. ScienceGuy
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    That would be the best answer ever...

    • one year ago
  44. ScienceGuy
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    Does this stuff make more sense to you now?

    • one year ago
  45. Gabbruce
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    Yes much more sense! I was mainly having trouble creating the set up but now I understand. Thanks

    • one year ago
  46. ScienceGuy
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    Not a problem at all. Glad I could be of help.

    • one year ago
  47. Gabbruce
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    Just one last question. Would it be 26.8J or no J?

    • one year ago
  48. ScienceGuy
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    The correct units would be degrees Celsius. This is because we are talking about temperature. Heat (Q) has units of J (joules)

    • one year ago
  49. ScienceGuy
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    Heat is energy, temperature is a relative measure of that energy so they are different things.

    • one year ago
  50. Gabbruce
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    Okay makes sense thanks again!

    • one year ago
  51. ScienceGuy
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    No problem. Later.

    • one year ago
  52. Gabbruce
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    Later.

    • one year ago
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