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Help With Thermodynamics

  1. Mar 6, 2005 #1
    A 0.50 kg metal bolt is heated to an unknown initial temperature. It is then dropped into a beaker containing 0.15 kg of water with an initial temperature of 21.0 C. The bolt and the water then reach a final temperature of 25.0 C. If the metal has a specific heat capcity of 899 J/kg * C, find the initial temparature of the metal.

    I am not sure even where to start.
  2. jcsd
  3. Mar 6, 2005 #2
    start with the conservation of energy. The initial and final energies of the system should be the same.
  4. Mar 6, 2005 #3
    i know that much, but i dont know which equation to use
  5. Mar 6, 2005 #4
    1/2(m)(mu)^2=constant, i figure but how do i find temperature from that
  6. Mar 6, 2005 #5
    You're supposed to be able to derive equations as simple as that, not necessarily remember them.
  7. Mar 6, 2005 #6
    What the hell is that?! BTW, this thread should be in the Homework Help section.

    The transferred energy equals change in temp. of substance times its mass times its heat capacity (assuming, of course, that the heat capacity is given as energy per unit temp per unit mass). Energy lost or gained by the metal equals energy lost or gained by the water...
    Last edited: Mar 6, 2005
  8. Mar 6, 2005 #7
    or is it cmt of water = cmt of bolt?
  9. Mar 6, 2005 #8
    Now I am not completely sure as I have only studied this, so far, in chemistry but here goes.

    You can work out the Energy Transferred to the water:
    Energy Transferred / E = Specific Heat Capacity / c x Mass of Water / m x Change in Temperature / ΔT

    Specific Heat Capacity for water is [tex]4.17 Jg^{-1}K^{-1}[/tex]

    [tex]E = 4.17 Jg^{-1}K^{-1} \times 150g \times (25 - 21)[/tex]

    [tex]E = 625.5 \times 4 = 2502 J[/tex]

    From this you can then do the reverse for the bolt.

    From here you can probably find something that will help and give an answer or someone will come along and say I am wrong. Either way.

    The Bob (2004 ©)
    Last edited: Mar 6, 2005
  10. Mar 6, 2005 #9
    um, well the answer is 81 degrees celcius, if u solve for (25-x) in that equation ur given delta t, which should be change from some reference point in the problem which im not seeing
  11. Mar 6, 2005 #10
    The 2502J is the energy that the water transfers to its surroundings. This is the energy that the bolt must have given the water.

    The Bob (2004 ©)
  12. Mar 6, 2005 #11

    The problem with your first equation is that the (25-21) is in celcius and the 4.17 j * g^-1 * k^ -1 is in kelvin.
  13. Mar 6, 2005 #12
    er never mind taht doesnt matter
  14. Mar 6, 2005 #13
    i dont know i cant get the answer ive tried a lot of things
  15. Mar 6, 2005 #14
    I was going to say. This really doesn't matter so long as the units are the same (and not farenheit :wink:).

    The Bob (2004 )
  16. Mar 6, 2005 #15
    Personally I feel the heat capacity of the bolt is too high but that might just be me.

    The Bob (2004 ©)
  17. Mar 6, 2005 #16
    i agree thats what i was getting at
  18. Mar 6, 2005 #17
    I think I might be close to cracking it. The number you gave was [tex]899 J kg^{-1}[/tex]

    I needed it in [tex]Jg^{-1}[/tex]

    This means that I need to convert it. This means that [tex]899 J kg^{-1} = 899000Jg^{-1}[/tex]

    The Bob (2004 ©)
    Last edited: Mar 6, 2005
  19. Mar 6, 2005 #18
    The amount of energy needed to increase the temperature of the water by 4°C is going to be what has already been worked out:

    [tex]E = (4.17 Jg^{-1} K^{-1})(150g)(25-21) = 2502J[/tex]

    So this is the energy give to the water. This is also the energy give up by the metal bolt. So:

    [tex]2502J = (899 Jkg^{-1}K^{-1})(0.5kg)(T_i - 25)[/tex] when [tex]T_i[/tex] is the inital temperature.

    I got most of the help from here. See if you can interprete it.

    I think the specific heat capacity is wrong. I think it should be 0.08.

    The Bob (2004 ©)
    Last edited: Mar 6, 2005
  20. Mar 6, 2005 #19
    hmm, i see, im really close to getting it, i think u have to convert the top to kg as well
  21. Mar 6, 2005 #20
    im not sure but if my memory serves me right energy is hand in hand with kg, not g.
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