A sample of water with a mass of 762.8 g and an initial temperature of 14.61 gains 119200 Joules. What is the final temperature?

Answer 1

q = 119200 J

m = 762.8 g

C = 4.184 J/gº

∆T = ?

q = mC∆T

∆T = q/mC = 119200 J/(762.8 g)(4.184 J/gº) = 37.35º

Final temperature = 14.61º + 37.35º = 51.96º