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Temperature and Thermal Energy

  1. Apr 2, 2009 #1
    Before I begin, Please note that for some reason I don't really grasp physic equations which is odd since I had a A in chemistry but physics just confuses me and also, I am taking this class to prepare me for college. My pencil broke and I didn't think we would have homework of the stuff we learned so quick, so I decided not to ask anyone for a pencil and just listened through the rest of the lecture but now I regret it.

    1. A 4.00 x 10²-g glass coffee cup is at room temperature, 20.0°C. It is then plunged into hot dishwater, 80.0°C. If the temperature of the cup reaches that of the dishwater, how much heat does the cup absorb? Assume the mass of the dishwater is large enough so its temperature doesn't change appreciably.



    2. Relevant equations
    The only thing I can think of is Q=mC∆T, but im pretty sure that is the wrong formula.


    3. The attempt at a solution

    m = 4.00 x 10²g = 4000 / 1000 = 4kg.
    Ti = 20.0°C
    Tf = 80.0°C
    C = 4180

    Q = (4)(4180)(80-20)
    Q = 1,003,200 j/kg * K
     
  2. jcsd
  3. Apr 2, 2009 #2

    rl.bhat

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    m = 4.00 x 10²g = 4000 / 1000 = 4kg.
    Check this one.
     
  4. Apr 2, 2009 #3
    oh, um the 1000 is because i needed to convert the grams to kilograms.
     
  5. Apr 2, 2009 #4

    rl.bhat

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    It should be 400/1000 kg
     
  6. Apr 2, 2009 #5
    oh yes lol, I have a habit of entering in my calc the x10....

    Q = (0.4)(4180)(80-20)
    Q = 100,320 J/Kg * k
     
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