1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Thermo physics question

  1. Jan 22, 2004 #1
    A 1.0 g lead bullet at 33°C is fired at a speed of 250 m/s into a large block of ice at 0°C, in which it becomes embedded. What quantity of ice melts?


    Heres what I have:

    From ΔQ = mcΔT,

    mA = dQ/[c(dT)]

    And from ΔQA = -ΔQB

    ...

    Further cogitations are in process. Meanwhile please respond, thanks.
     
  2. jcsd
  3. Jan 22, 2004 #2
    Re: Thermo

    1. The kinetic energy in the bullet is changed to internal energy (increase in temperature) when it stops inside the ice block. You can use the following formula to calculate the change in temperature of the bullet.
    1/2 mv2 = mcLΔT
    So the final temperatre of the bullet when embeded inside the ice block = 33 + ΔT

    2. Assume the temperature in the ice block stays at 0°C.
    The final temperature of the bullet will be 0°C, which means the energy released when the bullet is cooled from (33 + ΔT)°C to 0°C is used to melt the ice block.
    So you can use the following forumla to find out the quantity of ice melted.
    mLcL(33 + ΔT) = miceLice

    Where
    mL = mass of the lead bullet
    cL = specific heat capacity of the bullet
    mice = mass of ice melted
    Lice = latent heat of fusion of ice
     
    Last edited: Jan 23, 2004
  4. Jan 24, 2004 #3
    Re: Re: Thermo


    How is the Final Temperature of the Lead bullet 33 + ΔT?

    Perhaps you made a typo?

    By solving the KE = Q equation, i found Tf = (v^2/2c) + 33Celsius.
     
  5. Jan 24, 2004 #4
    93.781437125748502994011976047904 grams is the answer i got. not sure if i did it right or not.
     
  6. Jan 30, 2004 #5
    ok the answer is 0.107g...

    can someone do this problem with another approach pls?
     
  7. Jan 30, 2004 #6

    ShawnD

    User Avatar
    Science Advisor

    Re: Thermo

    All the changes in energy add up to 0 so base an equation around that using all changes in energy.

    [tex]mc \Delta T + mL_v - \frac{1}{2}mv^2 = 0[/tex]

    Check your signs. Terms gaining energy should be positive, losing energy should be negative. The velocity part is a loss in energy so that term is negative. Changes in temperature will always work themselves out with the proper sign; just be careful of latent heats and kinetic energy.

    mcT is the bullet's temperature change, mLv is the ice melting and 1/2mv^2 is the kinetic energy of the bullet. From there just start filling in the equation. The only variable is the mass of ice melting.
     
    Last edited: Jan 30, 2004
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Thermo physics question
  1. Thermo Question (Replies: 6)

  2. Thermo Physics Questions (Replies: 11)

  3. Thermo question (Replies: 1)

  4. Thermo Questions (Replies: 1)

Loading...