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Temperature problem

  1. Feb 28, 2005 #1
    :confused:

    The formation of condensation on a cold glass of water will cause it to warm up faster than it would have otherwise. If 4.5 gm of water condenses on a 122 gm glass containing 222 gm of water at 5.0 deg C, what will the final temperature be? Ignore the effect of the surroundings.

    The answer for this is 14.9 deg C

    But...I can't figure out how to get that answer. lol

    Here is what I am doing, where do I go wrong?

    (1 cal/g deg C)(222g)(Tf-5 degC) + (.2 cal/g deg C)(122g) (Tf-5 deg C)
    (122cal/ deg C)(Tf-5 deg C) + (24.4 cal/deg C)(Tf-5 deg C)
    (540 cal/g)=(146.4 cal/ deg C)(Tf-5 deg C)
    /146.4 cal/ deg C
    3.7=(Tf-5 deg C)
    8.7=Tf
     
  2. jcsd
  3. Feb 28, 2005 #2
    Trial and error. :)
    I figured out how to work it.

    Q=(4.5g)(540cal/g deg C)=2430 cal
    (1 cal/g deg C)(222g)(Tf-5 degC) + (.2 cal/g deg C)(122g) (Tf-5 deg C)
    (122cal/ deg C)(Tf-5 deg C) + (24.4 cal/deg C)(Tf-5 deg C)=246.4 cal/ deg c(tf-5 deg C)
    2430 cal/246.4 cal/deg c=14.9 deg C
     
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