# Another calorimeter problem, and %error

1. May 3, 2007

### erdfcvtyghbn

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. May 3, 2007

### Mindscrape

The percent error will be the discrepancy between the experimental value and the known value over the known value.
$$percenterror = \frac{|x_{exp} - x_{known}|}{x_{known}}*100$$

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