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Specific Heat of aluminum Problem

  1. Nov 16, 2005 #1
    "A piece of aluminum (specific heat 910 J/kg-C) of mass 200g at 80.0 deg Celcius is dropped into a styrofoam cup filled with 100mL of water at 20.0 deg C. What are the final temperatures of the water and the aluminum?"

    This is what I got:

    I figured out the amount of heat energy transferred for each using the equation Q = mc(delta T)

    m = 200g = .200kg
    c = 910 J/kg-C)
    T = 80.0 deg C

    Q = (.200kg)(910 J/kg-C)(80.0C)
    Q = 14560 J

    m = 100mL = 100g = .100kg
    c = 4190 J/kg-C)
    T = 20.0 deg C

    Q = (.100kg)(4190 J/kg-C)(20.0C)
    Q = 8380 J

    Ok, now this is where I'm confused. I don't know how to incorporate both of the temperatures when they are interacting with each other, or even what I did above was necessary.

    Any suggestions? Thanks
  2. jcsd
  3. Nov 16, 2005 #2
    Try looking at it again from the beginning.

    Q = (integral from Ti to Tf)mcdT
    What does this mean? What are you interested in finding and how can this help you? I'll tell you that what you've done for calculating heat won't be very helpful. So scrap that, you have the right idea to look at the heats, but not quite like that.
    (what you did for the aluminum heat was Q = (integral)mcdT where dT = 80C, which means that (Tf-Ti) = 80. But here you want to find a Tf from an intial T that you know is 80C)

    Since the water is in a styrofoam cup, we could assume there is no heat loss to the surroundings. What will happen after a period of time to the temperatures of the aluminum and water when they are in contact? Here's a hint, they eventually reach thermal equilibrium with each other.
    Ok, next.. how can we relate the heat transfer of the water(Qwater) to the heat transfer of the aluminum(Qal)? What is Qal and Qwater? Well,

    Qwater = (integral)mc(water)dT
    Qal = (integral)mc(al)dT

    Remember, the aluminum is going from temperature 80 to some Tf, and water going from temperature 20C to some Tf too.

    Now, if i have body A and i place it in thermal equilibrium with body B and heat is transfered from one body to the other(that is to say that their temperatures are not equal to start with), then the heat transfered into body A is related to the heat transfered from out of body B how?
    Qa = -Qb. Right? Do you understand why?

    So, relate Qwater to Qal, then solve for your temperature(When you relate Qal to Qwater, you will have only 1 unkown!)

    Well, i hope this makes sense to you and will help you! If you're still lost or confused by how i put things, just let me know and i'll try to word my thoughts better! :D
    Last edited: Nov 16, 2005
  4. Nov 26, 2005 #3
    Sorry this reply took so long, and thanks for your previous post.

    For Al, Ti = 80C
    For Water, Ti = 20C

    So I should use this equation? Qwater = mc(water)d(Tf-Ti) where I know m, c, and Ti

    Qal = mc(al)d(Tf-Ti) where I know m, c, and Ti again

    But now I have two unknowns and I'm stuck once again.

    Any suggestions? :confused: Thanks
  5. Nov 26, 2005 #4
    "Qa = -Qb. Right?"

    So if Qal = mc(al)d(Tf-Ti) and Qwater = mc(water)d(Tf-Ti) and Qal = -Qwater, then would this be the correct equation with only one unknown?::

    mc(al)d(Tf-Ti) = -[mc(water)d(Tf-Ti)] ?

    Where I know both m, both c, and both Ti ?

    Then I can solve for Tf which should be equal to each other because of thermal equilibrium, correct?
  6. Nov 27, 2005 #5
    Am I in the right ballpark with what I have written above?
  7. Nov 27, 2005 #6


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    Homework Helper

    yes, they end up with the same Temperature.
    (Otherwise the answers depend on how long the Aluminum is left in the water!)
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