Another calorimeter problem, and %error

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Homework Statement

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.

Homework 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

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 I am not sure if its right, and I am not sure how to find percent error

 

[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