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Homework Help: Specific Heat and Phase Change/Latent Heat Problems PLEASE HELP

  1. Nov 20, 2005 #1

    I have a couple problems that I have tried but I'm not sure if I worked them properly. Any advice or help is appreciated.

    One problem is: An electric immersion heater has a power rating of 1500W. If the heater is placed in a liter of water at 20 degrees C, how many minutes will it take to bring the water to a boil? Assum no heat loss except to the water.

    I tried it this way. Q-Pxt, P=1500W Q=mcDeltaT=1kg(4186)(-80 (change in temp))=334880. Then I took 334880 and divided by 1500 to get t... Is this the right way to do it or do I need to use a different equation?

    The other isA volume of .5 L water at 16 deg. C is put into an aluminum ice tray of mass .25 kg at the same temp. How much energy must be removed from this system by the refrigerator to turn the water to ice at -8 deg. C?

    I tried it by saying .5L=.5kg, Qgained=.25kg(920)(-8-16)=-5520
    Qlost=.5(4186)(-24)=-50232, dividing these gives -.11 J... I am sure this is not the correct way. If someone could steer me in the right direction, I'd appreciate it...

    Also, I do better with numerical explanations then written ones. What may sound normal to you might be over my head or sound completely different. Not trying to be weird, but it's true! And if anyone checks older posts, if there are any I haven't responded to in the past, it is only because I didn't get a response for a long time and did it on my own....

    Please help me with these. I'd liek to know what I'm doing here...
  2. jcsd
  3. Nov 20, 2005 #2


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    Homework Helper

    in the first one, the temp change is +80 K ... keep track of signs!

    In the second, you need to remove Energy from the *water* and the metal.
    First take out enough E to make liquid water @ 273 K
    then take out E to make solid water @ 273 K
    then take out E to make it solid water @ 265 K .

    Part of the subject (any subject) is the vocabulary ...
    another part is the preferred phrasing and syntax.
    You have to learn to USE these to communicate with.
    Last edited: Nov 20, 2005
  4. Nov 20, 2005 #3
    So I did the first question correctly?

    In the second, you need to remove Energy from the *water* and the metal.
    First take out enough E to make liquid water @ 273 K
    then take out E to make solid water @ 273 K
    then take out E to make it solid water @ 265 K .

    I understand this, I am just not sure whether to apply Q=mcdeltaT or Q=mL and to which part... I am confused how to apply these concepts.

    I know that the vocab is important, I was just saying that sometimes it is difficult to understand some things when they are written on a casual message board where people don't necesarily speak the clearest...

    I have another problem. The wall of a house is composed of solid concrete with outside brick veneer and is faced on the inside with fiberboard. The fiberboard width is 2 cm, the concrete width is 15 cm, and the brick width is 7 cm. If the outside T is -10 degrees, and the inside T is 20 degrees, how much E is conducted through the wall with dimensions 3.5mx5m in 1 hour?

    I want to try something along the lines of k1A(T-T1)/d1=k2A(T2-T1)/d2 or something along those lines, but I know I have to use a T3, but I'm not sure how to find the third T. If you or anyone could help me set that up, that would be great.
  5. Nov 20, 2005 #4
    You need to listen to lightgrav,

    You must do the problem in 3 steps (+1 for the Al.)

    Take the enrgy lost. (You do this correctly) but do it three times just as lightgrav says but you can use Celcius instead of Kelvin. Use Q=mL only for the phase change Water to Ice.

    Then you must do 1 extra step for the energy for the tray.

    Add everything up and that will be the total Q you are looking for.

  6. Nov 21, 2005 #5
    Ahh, I see it now,

  7. May 8, 2009 #6
    Aliminum is not changing its state your calculation is correct.
    Water is changing its state.Total heat energy is losted by water will be addition of heat lost to cool 0 C, Heat lost during melting and heat lost to cool -8 C.
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