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Temperature & Heat

  1. Jul 6, 2008 #1
    1. The problem statement, all variables and given/known data

    A thermos contains 155 cm3 at 87 degrees C. To cool the coffee, you drop two 12.2-g ice cubes into the thermos. The ice cubes are initially at 0 degrees C and melt completely. What is the final temperature of the coffee in degrees Celsius? Treat the coffee as if it were water

    2. Relevant equations

    (m*Lf)ice= (c*m*Delta T)water

    3. The attempt at a solution

    (.0244)(33.5x10^4)= (4186)(155)(Tf-87.0)

    T f= 86.987 degrees C
     
    Last edited: Jul 6, 2008
  2. jcsd
  3. Jul 6, 2008 #2

    Hootenanny

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    What happens to the water that results from the ice melting?
     
  4. Jul 6, 2008 #3
    When the ice melts, it cools the coffee so the overall temperature decreases.
     
  5. Jul 6, 2008 #4

    Hootenanny

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    Correct. Does the temperature of the melted ice stay at 0oC or does it change?
     
  6. Jul 6, 2008 #5
    The ice gains heat and the coffee loses heat
     
  7. Jul 7, 2008 #6

    Hootenanny

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    Correct, but you haven't taken this into account in your calculations have you? You need to take into account the temperature change of the ice-water after it has melted.
     
  8. Jul 28, 2008 #7
    i think after ice melt , temperature of water (for Ice melt) will up and water in thermos will drop

    SO

    (m*Lf)ice + (m*c*(Tf-0))ice = (c*m*Delta T)water

    (.0244)(33.5x10^4)+(.0244)(4186)(Tf-0)= (4186)(155)(87-Tf)
     
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