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March 30, 2017

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How would i find the following? I just need an explanation. I'm stuck. I have many other questions like this and i need to know how to do them. The specfic heat of ice is 2.05 J/g.



In a calorimeter, 10.0 g of ice melts at 0oC. The enthalpy of fusion of the ice is 334 J/g. How much heat was absorbed?

0.33 kJ

3.34 kJ

33.4 kJ

334 kJ

  • Will Someone please respond! Chem - ,

    q = mass ice x heat fusion = ? If you have others of a similar nature, they are

    At a phase change,
    mass x heat fusion at melting point OR
    mass x heat vaporization at boiling point.l

    WITHIN a phase,
    q = mass x specific heat in that hase x (Tfinal-Tinitial)

  • Will Someone please respond! Chem - ,

    The answer would be the last choice, but what strikes me is when i did the other problem i got the same answer. The other problem is
    In a calorimeter, 10 g of ice absorbs heat with an enthalpy of fusion of 334 J/g. What is the heat absorbed? q = mCHf

    6.68 J

    334 J<---

    3.34 kJ

    6.68 kJ

  • Will Someone please respond! Chem - ,

    Will you please help me with this also?

    How much heat do you need to raise the temperature of 150 g of ice from 30oC to 15oC?
    4.61 kJ <--This is my answer.Is itright?

    4.68 kJ

    5.49 kJ

    9.40 kJ

  • Will Someone please respond! Chem - ,

    What is missing from these problems is the temperature of the ice. If it is at zero C, then the heat absorbed to melt it AND leave the temperature at zero C is

    10g x 334 J/g = 3340 J = 3.34 kJ.
    I suspect that since the specific heat of the ice is given in the problem that you didn't type all of the problem and that the ice was not at zero C in one of them.

    The second problem is the same. If it is at zero C

  • Will Someone please respond! Chem - ,

    I think you made a typo for the 150 g problem. You must have meant ice at -30 going to -15. Then use the "within" formula I showed you at the beginning.
    q = mass x specific heat x (Tf-Ti)
    q = 150 x 4.184 x [(-15-(-30)]=
    q = 150 x 4.184 x 15 = 9414 J = 9.41 kJ (You get 9.40 kJ if you use 4.18 J/g*C for specific heat.)

  • Will Someone please respond! Chem - ,

    No there is no temperature for the second one in the question, but there is a specific heat for ice in a data table. It is 2.05 J/g.

    Yes that is what i meant they are both negatives.

  • Note! NOTE annonymous. - ,

    I made a horrible mistake. I used the specific heat of water when the problem clearly says it is ice going from -30 to -15. It should be
    q = mass x specific heat x (Tf-Ti_
    q = 150 x 2.05 J/g*C x (15)
    q = 4.61 kJ and your answer is correct.

  • Will Someone please respond! Chem - ,

    Thank you sir for correcting it. I hope you read my other response!

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