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818jasmine

  • 3 years ago

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?

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  1. Owlfred
    • 3 years ago
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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. Ms.bio
    • 3 years ago
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    wow, this one is thermodynamics right?

  3. 818jasmine
    • 3 years ago
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    yes

  4. 818jasmine
    • 3 years ago
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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.

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

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

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

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

  9. rsvitale
    • 3 years ago
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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)

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

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

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

  13. rsvitale
    • 3 years ago
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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.

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

  15. rsvitale
    • 3 years ago
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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.

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

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

  18. rsvitale
    • 3 years ago
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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

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

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

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

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

  23. rsvitale
    • 3 years ago
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    yeah

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

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

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

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

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

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

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

  31. 818jasmine
    • 3 years ago
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    i have 0.2619

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

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

  34. 818jasmine
    • 3 years ago
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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

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

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

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

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

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

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