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Specific heat capacity question

  1. Nov 30, 2011 #1
    1. The problem statement, all variables and given/known data
    1. When a car brakes, an amount of thermal energy equal to 112500J is generated in the brake drums. If the mass of the brake drums is 28 kg and their specific heat capacity is 460.5 Jkg^-1K^-1?
    2. A piece of iron of mass 200g and temperature 300°C is dropped into 1.00kg of water of temperature 20°C. What will be the eventual temperature of the water? Specific Heat Capacity of iron is 450Jkg^-1 K^-1, water 4200Jkg^-1 K^-1

    2. Relevant equations
    • Q=m c delta t
    • Q=c detla t

    3. The attempt at a solution
    1. (112500/28)/460.5 = 8.7 °C
    2. (4200+450)/(300+20)=14.53°C


    I am 99% sure that I got the first one right, however I don't think the second one is right because no one knew exactly how to do that one from my class. Could you guys steer me in the right direction :) ?
     
    Last edited: Nov 30, 2011
  2. jcsd
  3. Nov 30, 2011 #2
    re 2 your work is not correct.

    use conservation of energy

    energy given out by iron = energy gained by water
     
  4. Nov 30, 2011 #3
    You have misprinted 1125000 as 112500 in your calculation
     
  5. Nov 30, 2011 #4
    @technician Oh, sorry 'bout that.
    @grzz Don't I like need to know the final temperature of water to calculate this? If not, which equation should I use?
     
  6. Nov 30, 2011 #5
    You do not know the final temp of the water but you do know that the iron and water do end up at the same temp.
    write an expression for the heat energy given up by the iron and an equation for the heat energy gained by the water.
    These should be equal (ignoring heat losses) and you should find the only unknown quantity is the final temperature.... (call it ∅)
    This is using conservation of energy
     
  7. Nov 30, 2011 #6
    And lastly, whenever you do problems of this nature where two things with differing temperatures are mixed (no chemical reaction of course), the final temperature is ALWAYS somewhere between the temperatures of the two mixing constituents..
     
  8. Nov 30, 2011 #7
    What about this?
    Q=(4200+450)/((0.2x300)+20)

    Which makes it 58.125°C
    Is this correct?
     
  9. Nov 30, 2011 #8
    Use H = m x C x Δ∅ and put the numbers in for the iron
    Then use the same expression with the appropriate numbers for water ....Δ∅ ... is the difficult bit !!!!
    Make these equal and see if you can get the final temp....∅
     
  10. Nov 30, 2011 #9
    Would it be something around 36°C? I really can't find the change in temperature part in the equation...
     
  11. Dec 1, 2011 #10
    Let Tmix be the final temperature of the mixture.

    Heat lost by the iron = massiron * specific heat iron * (Tinitial - Tmix)
    Heat gained by water = masswater * specific heat water * (Tmix - Twater)

    Equate the above and solve for Tmix.
     
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