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Bullet stopped by sandbag: heat and temperature changes

  1. Jul 23, 2007 #1
    1. The problem statement, all variables and given/known data

    1. A bullet travelling at a speed of 360 m s-1 is stopped by a sandbag. Assuming half of the energy of the bullet becomes heat energy that is absorbed by the bullet, calculate the increase in temperature of the bullet. (Specific heat capacity of bullet = 160 J kg-1 C-1)

    2. A lead ball of mass 320 g is dropped from a height of 12 m. The collision between the ball and the ground is completely inelastic. Assuming all the energy of the ball goes into heating it, calculate the change in temperature of the ball. (Specific heat capacity of lead = 128 J kg-1 C)

    3. Water has a high specific heat capacity. In a water heater system in the bathroom, the flowing water has to be heated up in a short period of time. IN a certain water heater, 1.2kg of water at an initial temperature of 25 celcius passes through the heating coil every 20 x. What is the power heater if the heating system delivers water at a temperature of 40 celcius? (specific heat capacity of water = 4200 J kg-1 C)

    2. Relevant equations


    3. The attempt at a solution

    Im currently on a revision and i hope someone could help me out with these questions. Guide, steps, and explanation is greatly appreciated.
  2. jcsd
  3. Jul 23, 2007 #2


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    Specific heat capacity is the amount of energy in joules needed to raise 1kg of the substance 1deg C.
    1, The kinetic energy of a moving object (in joules) is 0.5 x mass in kg x velocity (m/s) * velocity(m/s)
    2, The potential energy of a body (in joules) is height(m) x mass(kg) x g where g is 9.8m/s/s
  4. Jul 23, 2007 #3
    ohw..thnx for the formulas...i'll try to do first. ^^
  5. Jul 24, 2007 #4
    can anyone correct me if i am wrong.

    question 1
    0.5 x 360 = 160 x 0(teta)
    0(teta) = 1.125 J
    *why do we use 0.5
    *is my answer correct? pls do tell me if im wrong

    question 2
    12 x 0.32 x 9.8 = 0.32 x 128 x 0(teta)
    37.632 = 40.96 x teta
    teta = 0.91875 J
    * is this correct?

    question 3
    no idea at all. Pls do teach me how to solve it
  6. Jul 24, 2007 #5


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    This is incorrect. You should write down the equations you need to use first, instead of trying to patch one together.

    Firstly, you are told that the bullet is travelling at 360m/s. So, what is the kinetic energy of the bullet, using the formula [itex]E=1/2\cdot mv^2[/itex]? You are then told that half of this is converted into heat energy. So, you want to find the change in temperature that E/2 gives to the bullet. Rearrange [itex]Q=mc\Delta\theta[/itex] and solve for delta theta.

    This is right, presuming you've performed the arithmetic correctly. In future, though, it would be helpful to write down what equations you were using, and justification for using them. For example, you say that the kinetic energy as the ball hits the ground is equal to the potential energy of the ball when it is dropped. This gets converted into heat energy which raises the temperature of the ball by the amount given.
  7. Jul 24, 2007 #6
    wow, thnx cristo, i've found the solution to question 1. Okay, i'll try to write down the formulas.

    question 3
    i assume we'll be using the formula Pt=mc[tex]\theta[/tex]
    so my solution,
    P(20s) = 1.2 x 4200 x (40-25)
    P(20) = 75600
    P = 3780 W

    is this right?
  8. Jul 24, 2007 #7


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    I'm not sure to be honest. I don't know about the assumption that it takes a second for the water to circulate the heating coil, but then, I can't see another way for the question to be done.

    Perhaps someone else will come along and be able to clarify.
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