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Ice melted by impact of lead bullet

  1. May 8, 2016 #1
    1. The problem statement, all variables and given/known data
    A 3.20 g lead bullet at 28.0°C is fired at a speed of 210 m/s into a large block of ice at 0°C, in which it becomes embedded. What mass of ice melts (g)?



    2. Relevant equations
    I thought that the KE of the bullet would be transferred into heat


    3. The attempt at a solution
    I used MV^2/2 = 0.0032x210^2/2
    =70.56J
    divided that by heat of fusion of ice, 80cal/g
    =0.882g
    Its wrong tho
    someone pls help :) thanks.
     
  2. jcsd
  3. May 8, 2016 #2

    SteamKing

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    So you're just dividing numbers willy-nilly and disregarding units? Are calories the same as joules?
     
  4. May 8, 2016 #3
    • Poster has been reminded not to use text speak at the PF
    Are you talking about where i divided 70.56J by 80cal/g because i also tried converting 70.56J into cals and dividing it by 80cal/g again and it didnt work. If not please expand on what you mean buddy
     
    Last edited by a moderator: May 9, 2016
  5. May 8, 2016 #4

    SteamKing

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    You didn't show any attempt at converting joules to calories or vice versa, so how are we to guess what you did? How do we know you even did the right conversion?

    BTW, text speak is not welcome at PF.
     
  6. May 8, 2016 #5

    billy_joule

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    Presumably you are given this information for a reason. Eg don't just consider the bullets kinetic energy, include it's thermal energy too.
     
  7. May 8, 2016 #6
    < Mentor Note -- text speak edited out of post >

    so thermal energy is
    Q=mcΔT
    which works out to be 14.336 J
    but how do i incorporate that into my working?
    My working so far:
    We= energy from bullet
    =1/2 mv^2 + mc(sub)pΔT
    mass in kg
    the C(sub)p being specific heat of lead = 0.160 j/g degrees celsius
    which comes out to be 70.57J
    energy is fully absorbed by the blocked of ice
    ∴ We=Mi(Hf)
    Mi = mass that melted
    Hf = heat of fusion of ice
    ∴Mi=We/Hf
    =211.97
    which is wrong
    Can you please help by explaining where im going wrong.
    thx
     
    Last edited by a moderator: May 9, 2016
  8. May 8, 2016 #7

    SammyS

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    Please, knock off the "text speak".

    The words are spelled, you, please and thanks .
     
  9. May 8, 2016 #8

    SteamKing

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    To recap:

    The total heat provided by the bullet is 70.57 J (KE) + 14.336 J (ΔT)

    You were given the heat of fusion of ice as 80 cal/g

    You must convert joules to calories or vice versa in order to calculate the amount of ice melted by the bullet.
    It's not clear which conversion you are using.

    Mi = 211.97 what? grams? kgs? Please show units on your results.
     
  10. May 9, 2016 #9

    gneill

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    I see that the value you've used for the specific heat of lead is 0.160 j/g/C, or 160 J/kg/C). Was this a given value in the problem or text that it came from? The reason I ask is that in the tables I've seen the value is generally given to be around 128 J/kg/C.
     
  11. May 9, 2016 #10
    We weren't given a specific heat of lead value, so I just googled it
     
  12. May 9, 2016 #11

    gneill

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