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!

Hello, thermodinamic problem

  1. Jul 22, 2004 #1
    i have an easy problem that i can't solve

    if you throuth 100g of ice at a tempeture of -15C in a glass with 200g of water at a tempeture of 25C.

    what it the final tempeture of the system? the system is isolated, no heap can get in neither get out.

    i try to solve this problem using specific heat capacity of water and ice, but i get weird result.
    for water i used heat capacity = 4190 J/kg*K
    heat capaciy of ice = 2220 J/Kg*K

    can any one give an idea on how to setup the equation?
  2. jcsd
  3. Jul 22, 2004 #2
    Have you considered the latent heat involved when the ice melts?
  4. Jul 22, 2004 #3
    Yes i consider that. but how do i know all the ice is melted?

    if all the ice is melted


    Q_1 = cm(T_i - T_f)
    Q_2 = Lm

    for the ice the total heat Q_1 + Q_2


    Q = cm(T_i - T_f)

    the final equation looks like

    Q_1 + Q_2 = Q

    then using the correct values of c,m and L i just solve for T_f

    is this correct??
  5. Jul 22, 2004 #4
    That looks about right but I would calculate the temperature difference as Tf -Ti. If you do it the other way round, a negative sign somewhere might lead you to the wrong answer.
  6. Jul 22, 2004 #5
    oh yes that was a typo i mean to write T_f - T_i
  7. Jul 23, 2004 #6
  8. Jul 23, 2004 #7
    Thanks all for your help, but i can't get the answer to my problem.
    i'm using this values

    T_i = -15C = 258.15K \\
    m_i = 100g = 0.1Kg \\
    c_i = 2220 \frac{J}{Kg*K} \\
    L_i = 333 \frac{KJ}{Kg}

    T_i = 25C = 298.15K \\
    m_w = 200g = 0.2Kg \\
    c_w = 4190 \frac{J}{Kg*K}

    ok now, the equation for the ice:

    Q_1 = c_im_i(T_f-T_i) = c_i_m_iT_f-c_im_iT_i \\
    Q_2 = L_im_i

    the equation for the water

    Q = c_wm_w(T_f-T_i) = c_wm_wT_f-c_wm_wT_i

    no heat can get in neither get out so to get the final tempeture i solve for T_f the next equation

    Q_1+Q_2 = Q \\
    c_i_m_iT_f-c_im_iT_i + L_im_i = c_wm_wT_f-c_wm_wT_i \\
    T_f = \frac{c_im_iT_i-L_im_i-c_wm_wT_i}{c_im_i-c_wm_w}\\
    i get [tex]$T_f = 366.62K$[/tex] wrong\\

    Could you tell me where is my mistake??
  9. Jul 23, 2004 #8
    Ok, there are 3 parts to the problem. The first part is from ice at -15 to ice at 0. Then there is ice at 0 to water at 0. Then, it's water at 0 with water at some other temp to be determined. I'm currently trying to work out the equations, will post when done.
    Last edited: Jul 23, 2004
  10. Jul 23, 2004 #9
    This seems to be a peculiar problem. My first guess is that you will end up with some ice and some water both at 0C. Is this an actual question from a book?

    If you work it out step by step (I would write down equations but I've never used that LaTeX thing.):

    First the ice takes heat from the water. It will require the ice 3330J to go from -15 to 0C.

    When that heat is taken from the water, the new water temperature is 21.03C.

    Now, if the ice is to completely melt, it will need 33300J of energy but the water at 21.03C can only supply 17620J of energy before it itself gets to 0C.

    Therefore you end up with some ice/water equilibrium at 0C.

    PS: My calculations are based on the specific heat capacities and latent heat that you have used.
  11. Jul 23, 2004 #10
    I work out that you end up with 47.09g of ice and 252.91g of water in equilibrium. Can anybody confirm this? It's been so long since I did heat calculations so I might have done silly mistakes somewhere.
  12. Jul 23, 2004 #11
    yes the answer on the back of the book is 0C.

    thanks for your post now i know how calculated if the ice is melted or not.
    i dont know why i didn't think of it before.

    now i'll try to write down my equations
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?