J.R. S. answered 02/21/23
Ph.D. University Professor with 10+ years Tutoring Experience
Best to approach this problem in a step-wise manner.
Step 1: melt 60 g ice @ 0º. q = m∆Hfus = (60.0 g )(80 cal / g) = 4800 cal
Step 2: raise temp of 60 g of liquid from 0º to 65º. q = mC∆T = (60.0 g)(1.00 cal/gº)(65º) = 3900 cal
(note: the SH should be 1.00 cal/gº and not cal/g)
Step 3: Sum the calories and convert to kcal. 4800 cal + 3900 cal = 8700 cal x 1 kcal/1000 cal = 8.70 kcal