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Temperature/Heat Help

  1. Apr 15, 2008 #1
    I have two questions that I need help with. Any help will be appreciated.

    1. A lead bullet initially at 17oC just melts upon striking a target. Assuming that all of the initial kinetic energy of the bullet goes into the internal energy of the bullet to raise its temperature and melt it, calculate the speed of the bullet upon impact

    I know KE = 1/2 mv^2 and Q= mcdeltaT
    In this case, Q= 1/2mv^2
    so that mcdeltaT = 1/2mv^2
    This way the masses cancel and so cdeltaT = 1/2v^2 and the c of lead is 128.
    I know the initial temperature is 17 degrees, but what is the final temperature? The wording of the question is confusing me and I don't know what delta T is. It seems like such a trivial mistake.

    2. A cube of ice is taken from the freezer at -10oC and placed in a 110 g aluminum calorimeter filled with 280 g of water at a temperature of 25oC. The final situation is observed to be all water at 11oC. What was the mass of the ice cube?(cwater=4186 J/kg.oC, cal=900 J/kg.oC, cice=2000 J/kg.oC, Lf=33.5x104 J/kg)

    I know Q gained = Q lost
    The aluminum in this case is providing the energy, thus it is losing the heat. The ice is gaining heat.
    McdeltaT
    Mice*Cice(10) * MiceLf + MiceCw(25)= MiceCw(25-11) + MalCal(25-11)
    I think I'm on the right track, but after I plug in the values that I know and I solve for the mass of ice, I'm getting a very large number which means I'm making a mistake somewhere. Any suggestions?
     
  2. jcsd
  3. Apr 15, 2008 #2
    1. Final temperature is the melting point of lead. You will also have to take into account the latent heat of fusion:

    mc delta T + mL = 1/2 mv^2, where delta T = (m.p of lead) - 170C

    2. Q gained = [M(ice) * C(ice) * 10] + [M(ice) * L(f, ice)] + [M(ice) * C(w) * 11 ]

    Q Lost = [M(al) * C(al) * (25 -11)] + [M(w) * C(w) * (25 -11)]

    RHS is wrong: The 1st term should be M(w) * C(w) * (25 - 11), where M(w) = 280g
    LHS is wrong: 3rd term should be M(ice) * C(w) * 11 i.e. heat gained by the just melted ice to reach 11 oC
     
  4. Apr 15, 2008 #3
    Thank you so much for your help on the first problem. I overlooked the fact that I had to take into account the phase change. However, I'm still confused on the second problem.

    The equation I originally had should be this:
    Mice*Cice(10) + MiceLf + MiceCw(25)= MiceCw(25-11) + MalCal(25-11). Can you clarify whats wrong please, I sorta didn't understand what you said.
     
  5. Apr 15, 2008 #4
    Thank you so much for your help on the first problem. I overlooked the fact that I had to take into account the phase change. However, I'm still confused on the second problem.

    The equation I originally had should be this:
    Mice*Cice(10) + MiceLf + MiceCw(25)= MiceCw(25-11) + MalCal(25-11). Can you clarify whats wrong please, I sorta didn't understand what you said.
     
  6. Apr 15, 2008 #5
    What loses heat:

    Q Lost = [M(al) * C(al) * (25 -11)] + [M(w) * C(w) * (25 -11)]

    1st term: Aluminium loses heat by cooling from 25 to 11 oC
    2nd term: It's given in the problem that there's 280g of water already present in the container. This loses heat to the ice as it cools from 25 to 11.

    Q gained = [M(ice) * C(ice) * 10] + [M(ice) * L(f, ice)] + [M(ice) * C(w) * 11 ]

    1st term: Heat required to get ice to 0 oC from -10
    2nd term: Needed to melt that ice at 0 oC
    3rd term: That melted ice (which is now water at 0 oC) needs to be heated to the final temp of 11 oC. Hence this term

    So basically there are 3 objects:
    The ice, unknown mass, which undergoes heat(10 to 0) + melt + heat(as water now, 0 to 11);
    The aluminium(110g) which undergoes cool(25 to 11);
    The water at the beginning in the container(280g), which undergoes cool(25 to 11)
     
  7. Apr 15, 2008 #6
    thank you so much. i have to read these problems more carefully.
     
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