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Final Temp of Clay

  1. Sep 16, 2007 #1
    1. The problem statement, all variables and given/known data
    On a pleasant fall day(temperature of 21.0 C) a lump of clay (with mass of .885kg) is thrown against the wall with speed of 38.0 m/s. the clay deforms as it sticks to the wall, noiselessly. assuming no heat escapes into the air, what will the final temp of the clay be.(clay starts at same temp as the air) Cclay=2555j/kgK


    2. Relevant equations
    Q=cm(change in Temp)T
    Q=change in KE
    KE=1/2mv(squared)
    1/2mv(squared)=cm times the (change in temp)

    3. The attempt at a solution

    1/2(.885kg)(38.0 m/s)squared=(2555j/kgK)(.885kg)*Change in temp of clay.
    (.4425kg)(1444)=(2261.175)(change in temp of clay)
    638.97=(2261.175)(change in temp of clay)
    .283 C=change in temp of clay

    final temp=.283+21.0
    final temp=21.28


    i did it three times but the answer doesnt seem right to me. if it loses no energy shouldnt all of it be transformed into heat.
    some when please tell me what im doing wrong.
     
  2. jcsd
  3. Sep 16, 2007 #2
    I don't immediately see the problem. It appears correct.
     
  4. Sep 16, 2007 #3
    thank you, i was trying to think of a different way. but this was the only way i could think of and i wanted to make sure this was right because there is a follow up question to this. again thanks
     
  5. Sep 16, 2007 #4
    hey suraj i got a different answer to that problem. i did what u did and i got 28 degree celcius but i did not get 21.28 as my final answer.
     
  6. Sep 16, 2007 #5
    how did you get 28 C
     
  7. Sep 16, 2007 #6

    dynamicsolo

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    Homework Helper

    The kinetic energy transformed into heat is 639 J. With the given specific heat and the fact that the lump is about 1 kg in mass, it's hard to see how the temperature change would be *larger* than 1 degree C...
     
  8. Sep 16, 2007 #7
    well this thing is due tuesday. i know that. and i got 28 by doing exactly what u did and then i added the 28 with 21 to get my final temperature the 28 is the change in temp.
     
  9. Sep 16, 2007 #8
    did you do some mathematical mistake because 28 rise in temp is huge. i thought this was due monday
     
  10. Sep 16, 2007 #9
    he said u got till tuesday. First i found the kinetic energy which came out to be 638.97 and then i found the change in temp by using the kinetic enerygy which is heat. because that is the reason why they gave the velocity. and i got 28 degrees.
     
  11. Sep 16, 2007 #10

    dynamicsolo

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    Homework Helper

    Be sure to check your units or calculator input: the change of 0.28 C comes from

    639 J / [(2555 J kg/K)(0.885 kg)] .
     
  12. Sep 16, 2007 #11
    yeah that's exactly how i did it to get the change in temp. because u can't do the change in temp without the heat.
     
  13. Sep 16, 2007 #12
    you use 1/2mv squared=cm delta t
     
  14. Sep 16, 2007 #13
    yup. that's what i did and when is this thing due do u know for sure i believe he said it's due tommorow.
     
  15. Sep 16, 2007 #14
    cm delta t should come out to 2261.175j/k delta t
     
  16. Sep 16, 2007 #15
    k my fault make sense yup i forgot about the decimal ur answer is right sry. how did u do the second one and is this thing do tommorow.
     
  17. Sep 16, 2007 #16
    u use the heat in=heat out formula. ill show you in 10 min after i get home. it is due tomorrow.
     
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