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Cooling Lead Shot

  1. Nov 2, 2007 #1
    1. The problem statement, all variables and given/known data
    5 kg of lead shot at 97.9 °C are poured into 5 kg of water at 26.0 °C. Find the final temperature of the mixture. Use cwater = 4187 [(J)/(kg·° C)] and clead = 128 [(J)/(kg·° C)].


    2. Relevant equations



    3. The attempt at a solution
    How do I start this?
     
  2. jcsd
  3. Nov 2, 2007 #2

    Astronuc

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    Staff: Mentor

    There are the initial energy content of the water and of the lead shot.

    The energy must transfer from hot (lead) to cold (water), i.e. the difference (change) in energy content of the lead, which is related to the temperature change, must equal the change in energy of the water.

    And ultimately, they have the same temperature.
     
  4. Nov 2, 2007 #3
    So can I use Q = mc * change in temperature?
     
  5. Nov 2, 2007 #4

    Astronuc

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    Staff: Mentor

    Yes - realizing that the lead and water will have the same, as yet to be determined, temperature T.
     
  6. Nov 2, 2007 #5
    Q = (Ml * Cl * CH T) + (Mw * Cw * CH T)
    Q = CH T * [(Ml * Cl) + (Mw * Cw)]
    Q = CH T (21575)
    Q = Tf - Ti * (21575)
    Tf = (97.9C + 273K) * (21575) = 8002167.5
    Was I supposed to subtract the lead and water? Cause thats obviously wrong hehe.
     
  7. Nov 2, 2007 #6
    Why am I getting such a huge number???
     
  8. Nov 4, 2007 #7
    Does anyone know how to do this?
     
  9. Nov 4, 2007 #8

    Astronuc

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    Staff: Mentor

    The lead starts at some temperature Th and cools to temperature T, which one is trying to find, so the temperature change (Th - T) is proportional to thermal energy lost. At the same time, the water is starting at some temperature Tc and heats to temperature T, and the change (T - Tc) is proportional to the heat gained, and that heat (thermal energy) comes from the lead.

    Assuming not heat is lost from the system |[itex]\Delta{Q}_{Pb}[/itex]| = |[itex]\Delta{Q}_{water}[/itex]|

    and [itex]\Delta{Q}[/itex] = [itex]mc_p\Delta{T}[/itex]
     
  10. Nov 4, 2007 #9
    Okay so my m is just the mass of the lead?
    Change in Q = (5kg) * (128J/kg*C) * Change in T (97.9 - T)
     
  11. Nov 4, 2007 #10

    Astronuc

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    Staff: Mentor

    That is fine for the lead.

    Now do the same for the water.
     
  12. Nov 4, 2007 #11
    Change in Qpb = (5kg) * (128J/kg*C) * Change in T (97.9 - T)
    Change in Qw = (5kg) * (4187J/kg*C) * Change in T (26 - T)
     
  13. Nov 4, 2007 #12
    so I set them equal to each other
     
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