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Dropping steel ball into water

  1. Apr 30, 2014 #1
    1. The problem statement, all variables and given/known data
    If a steel ball of mass 100g and 100°c is dropped into 1litre of water at 20°c, what is the temperature rise? What would happen if it were dropped into a mixture of ice and water at 0°c?


    2. Relevant equations
    Q=mcΔT
    Q=mlm where ml is latent heat of melting
    Q lost = Q gained



    3. The attempt at a solution
    Qsteel = (0.1) (420) (-80) = -3360J
    the steel ball loses 3360J so the water must have gained 3360J

    Qwater = 3360 = (1)(4200)ΔT

    ΔT = 3360/4200=0.8°C
    so the waters temperature has increased by 0.8 degrees.

    For the second part it didnt specify how much ice so i just chose to have 1 litre of water with 150g of ice at 0°c.

    Qsteel = (0.1)(420)(-100) = -4200J
    so the steel ball has lost 4200J of heat and the water ice mixture will gain 4200J of heat

    Qwater = Qice = mlm

    mice= Qwater/lm = 4200/3.35x105 = 0.0125kg

    So only 12.5g of ice would be melted and the temperature wouldnt rise at all as the energy has gone into melting the ice and there is still a significant amount left.


    Am I on the right track here? Thanks
     
  2. jcsd
  3. Apr 30, 2014 #2
    The first part was not done correctly. The steel ball will not cool all the way down to 20 C. Both the steel ball and the water will end up at the same temperature.

    Chet
     
  4. Apr 30, 2014 #3
    i thought i had the right idea but the numbers came out unusual

    (0.1)(420)(x-100) = (1)(4200)(x-20)

    42(x-100) = 4200(x-20)
    42x-4200 = 4200x -8400

    then from here i got

    (42x-4200x)-4200= -8400

    -4158-4200= -8400

    -4158x = 4200 - 8400

    in the end i ended up with 100/99 which cant be right.

    if i calculate it with

    (0.1)(420)(x-100) = (1)(4200)(20-x)

    42x-4200 = 84000-4200x
    42x+4200x = 84000+4200
    4242x=88200
    88200/4242 = 20.79

    instead, i get 20.79°c which makes more sense but i thought delta T was always final minus initial

    edit: this also means that the steel ball loses 3326.4J instead
     
  5. Apr 30, 2014 #4
    The following equation is incorrect: (0.1)(420)(x-100) = (1)(4200)(x-20)

    It should read: (0.1)(420)(100-x) = (1)(4200)(x-20)

    The heat lost by the ball is gained by the water. Also, if you cancel before you multiply, the algebra is much simpler:

    (100-x)=100 (x-20)

    Chet
     
  6. Apr 30, 2014 #5
    so now i have

    (0.1)(420)(100-x) = (1)(4200)(x-20)

    4200-42x = 4200x - 84000

    4200 + 84000 = 4242x

    88200/4242 = 20.79

    (0.1)(420)(79.2) = 3326.4J

    so same answer before but a positive 3326.4J as the answer making the temperature rise 0.8c . i was expecting it to be negative because the ball is losing energy.

    thanks
     
    Last edited: Apr 30, 2014
  7. Apr 30, 2014 #6
    The solution I get to your original equation is 19.2 C. You must have made a compensating mistake in the algebra somewhere.

    Also, the 3326 J is the amount of heat lost by the ball.
     
  8. Apr 30, 2014 #7
    I put the whole equation into wolfram alpha to double check and got the same answer too http://imgur.com/iatFmIJ

    For the second question if I had 500g water and 500g ice would I use 0.5 for my mass of water or does the ice make it 1kg?
     
  9. Apr 30, 2014 #8
    With all due respect to wolfram alpha, it can't be the same answer if you replace (x-100) with (100 - x).
    Neither. After the steel ball is dropped in, a small amount of ice would melt to form liquid water, but the temperature of the ice water mixture would not change. All the heat from the steel ball would go into melting the small amount of ice.

    chet
     
  10. May 1, 2014 #9
    You said to use
    (0.1)(420)(100-x) = (1)(4200)(x-20), if I use x-100 I get 19.2. Apologies for missing the point
     
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