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818jasmine

when a 140g piece of metal at 70 degree Celsius is placed in 220g of water at 0 degree Celsius, the metal is cooled, and the water is warmed and both come to a final temperature of 10 degree Celsius. What is the specific heat of the metal?

  • 2 years ago
  • 2 years ago

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  1. Owlfred
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

    • 2 years ago
  2. Ms.bio
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    wow, this one is thermodynamics right?

    • 2 years ago
  3. 818jasmine
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    yes

    • 2 years ago
  4. 818jasmine
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    its actually physics but it's considered to be in the math category so i need a lot of help with this.

    • 2 years ago
  5. rsvitale
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    The heat associated with a certain mass changing temperature is

    • 2 years ago
  6. rsvitale
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    m*c*(dT)

    • 2 years ago
  7. rsvitale
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    the energy change in the water has to equal the energy change in the metal

    • 2 years ago
  8. rsvitale
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    Qm=140*c*(70-10) Qw=220*4.186*(10-0)

    • 2 years ago
  9. rsvitale
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    you set the two equal to each other and solve for c. That equation will give units of Joule/(gram Celsius)

    • 2 years ago
  10. 818jasmine
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    so would i be solving the equation that you suggested i'm confused

    • 2 years ago
  11. rsvitale
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    Have you learned the heat/energy required to change the temperature of a substance?

    • 2 years ago
  12. 818jasmine
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    no i don't quite understand how to solve problems like these.

    • 2 years ago
  13. rsvitale
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    ok we have the equation Q=mc(dT) where Q is heat, m is mass of whatever is changing temperature, dT is the change in Temperature.

    • 2 years ago
  14. rsvitale
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    the magnitude of the heat has to be the same for both the metal and the water

    • 2 years ago
  15. rsvitale
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    so you plug the known values into the heat equation for water and metal and set them equal to each other to find the specific heat you are looking for.

    • 2 years ago
  16. rsvitale
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    The magnitude of the heat is the same in both due to conservation of energy

    • 2 years ago
  17. 818jasmine
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    so basically what would be the numbers in the equation to plug into the problem if is the same.

    • 2 years ago
  18. rsvitale
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    Your final equation would look like: 140*c*(70-10)=220*4.186*(10-0) 140*60*c=220*10*4.186

    • 2 years ago
  19. rsvitale
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    and you want to find c in J/gC

    • 2 years ago
  20. 818jasmine
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    so how would i find c

    • 2 years ago
  21. rsvitale
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    140*60*c=220*10*4.186 8400*c=9209.2 c=9209.2/(8400) c=1.1 J/gC

    • 2 years ago
  22. 818jasmine
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    1.0963 would be c

    • 2 years ago
  23. rsvitale
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    yeah

    • 2 years ago
  24. 818jasmine
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    so what units should i use for this problem?

    • 2 years ago
  25. rsvitale
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    that c is J/(gC) or Joules per gram degree Celsius

    • 2 years ago
  26. rsvitale
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    the energy in Joules it takes to change the temperature of 1 gram of metal 1 degree celsius

    • 2 years ago
  27. 818jasmine
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    so would this be the same as 1.0963 cal/g degree Celsius ?

    • 2 years ago
  28. rsvitale
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    no 1 cal= 4.186 Joules

    • 2 years ago
  29. rsvitale
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    if you want calories divide your answer by 4.186

    • 2 years ago
  30. 818jasmine
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    i don't think my answer is right

    • 2 years ago
  31. 818jasmine
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    i have 0.2619

    • 2 years ago
  32. rsvitale
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    yep I think that's good

    • 2 years ago
  33. rsvitale
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    why dont you think it is right

    • 2 years ago
  34. 818jasmine
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    i don't know because i think that my units wont match up with the right answer that i just put up there

    • 2 years ago
  35. rsvitale
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    im not sure what you mean

    • 2 years ago
  36. 818jasmine
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    like should it say 0.2619 cal/g degrees Celsius, does that sound right?

    • 2 years ago
  37. rsvitale
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    oh yeah those are the units for specific heat

    • 2 years ago
  38. 818jasmine
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    thank you so much I got it right thanks

    • 2 years ago
  39. rsvitale
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    you're welcome :)

    • 2 years ago
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