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Homework Help: Specific heat and temp - please check

  1. Nov 11, 2007 #1
    In an insulated container, 0.50 kg of water at 80 degrees C is mixed with 0.050 kg of ice at -5.0 degrees C. After a while, all the ice melts, leaving only the water. Find the final temperature T_f of the water. The freezing point of water is 0 degrees C.
    I know that the specific heat of water is 4190 J/(kg*K), the specific heat of ice is 2100 J(kg*K), and that the heat of fusion is 334000 J/kg
    I also know that the expression for the amount of heat absorbed by ice is 0.05*2100(0-(-5))
    The expression for the amount of heat released by water is (0.5)(4190)(T_f-80)
    The expression for heat of transformation is (334000*0.05)

    And then I set these Q's that I found to 0. And my answer that I get is 71.78 degrees C.

    But I have a feeling I am doing something wrong. Any feedback will be awesome.
  2. jcsd
  3. Nov 11, 2007 #2
    no one can help....?
  4. Nov 11, 2007 #3
    lol i did like a LOT of problems like these in ap chem last year.. but i forgot some of the formulas.. could u put up some formulas? please
    Also, u should post this on the chemistry/bio/other science forum
  5. Nov 11, 2007 #4
    oh by the way.. it says -->4190 J/(kg*K) convert it to kelvin
  6. Nov 11, 2007 #5
    To solve a calorimetry problem like this one, Qnet = 0
    In absence of phase changes, amount of heat Q required to change the temp of an object of mass m from Temp_i to Temp_f, we use Q=mc(T_f-T_i) where c=specific heat/
    Since we are dealing with a phase change in this prob, we must also take into consideration the amount of heat involved here which is Q=mL where L is the heat of transformation.
  7. Nov 11, 2007 #6
    ok i think i remember.. basically the heat released must equal to heat absorbed.. so what is the total heat absorbed? the heat of transformation + the heat it takes change the temperature of ice to 0.. right? so set your first + third equations = second
  8. Nov 11, 2007 #7
    and it doesn't matter whether i use kelvin or celsius. as long as i am consistent throughout the prob, it doesn't matter...
  9. Nov 11, 2007 #8
    thats true.. ok sorry.. but i'm pretty sure what i said earlier is correct. is that basically what u did?
  10. Nov 11, 2007 #9
    I get 88.22 which is IMPOSSIBLE. how the heck can the temp rise??? Oy.
  11. Nov 11, 2007 #10
    lol why did u quote your own statements? i'm getting it to be 72.5 did u get 88.22 using what i said?

    The reason u got u got 88.22 is because u added 8.22 instead of subtracting.. you have to realize that u have to put a negative before 4190.. the reason is due to its being released. i mean this negative positive is all relative, when this is consistent.

    The reason im getting 72.5 is because i'm dividing by .55 instead of .5... because im adding the mass of the ice..
    (I think thats what u do, it makes sense right?)
    Last edited: Nov 11, 2007
  12. Nov 11, 2007 #11
    I did 525+16700 = 2095(T_f-80) and got T=88.2

    i quoted the problem so i could refer to it when i type my response so that i don't have to scroll up so far^^
  13. Nov 11, 2007 #12


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    Homework Helper

    Why? This is a typical introductory physics thermodynamics question.

    Remember that the heat lost by the water is equal to the heat gained by the ice. Be careful of your signs.
  14. Nov 11, 2007 #13
    your answer doesn't make much sense tho... why would temperature only decrease by 1 .. what i did to get 71.77 the first time was make my second equation negative since heat is being released. but i'm not so sure now, you know?
  15. Nov 11, 2007 #14
    ya i'm sorry hage. Well i know that but i suggested that because he posted it around 5. but no one responded..
  16. Nov 11, 2007 #15
    ya i was plugging the numbers off, i missed a zero. but the reason i'm getting it to be 72.5 is because i'm adding the mass of the ice.

    --i edited my comment, read it again
  17. Nov 11, 2007 #16
    I have noOoOo clue how you are arriving at that number. can you spell it out for me? and I totally understand the posting this on chem forum thing cuz i remember having to do these kind of probs in chem way back when.
  18. Nov 11, 2007 #17
    0.05*2100(0-(-5)) + (334000*0.05)=-(.55)(4190)(tf-80) ... ok
    i have .55 because i added the mass of the ice.. now i can be wrong, but i don't think so.. because it makes sense. I used to be very good in chemistry class as well lol. The negative is there because its being released..
  19. Nov 11, 2007 #18
    I don't understand this "because I am adding the mass of the ice" thing. You added the mass 0.05kg to my answer of 71.77? Now that doesn't make much sense.
  20. Nov 11, 2007 #19
    no thats not what i'm saying
  21. Nov 11, 2007 #20
    lol you keep posting just when i have finished responding...cancel what i said above for now
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