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Homework Help: Another calorimeter problem, and %error

  1. May 3, 2007 #1
    1. The problem statement, all variables and given/known data
    calorimeter mass = 70 g
    specific heat of calorimeter = .1 cal/gC
    cal. and water mass = 200g
    temp of water and cal = 65 C
    mass of cal, water, and ice = 220 g
    temp of ice = 2 C
    final temp of cal, water, and ice = 30 C
    Find Lf for this ice.
    Find the percent error.

    2. Relevant equations
    m Lf + m c (change in temp) = m c (change in temp) + m c (change in temp)
    left of equals sign is for ice, 1st m c T on right is water, 2nd is calorimeter

    3. The attempt at a solution
    20 Lf + 20 2.09 28 = 130 4.19 -35 + 70 (.1 x 4.19) -35
    i got -1063.07 for Lf, but im not sure if its right, and im not sure how to find percent error
  2. jcsd
  3. May 3, 2007 #2
    The percent error will be the discrepancy between the experimental value and the known value over the known value.
    [tex]percenterror = \frac{|x_{exp} - x_{known}|}{x_{known}}*100[/tex]
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