Question

500 J of heat are added to 59 g of water initially at 20°C.
(a) How much energy is this in calories?

(b) What is the final temperature of the water?

I know that to find (a) I have to use the constant 4.19 J and divide it into the 500 J yielding 119.33 calories. How do I find final temperature?
1J =0.2388 cal
500 J = 119.4229 cal.
Q =mcΔT,
ΔT = Q/mc = 500/0.059•4180 =02.03 oC
T1 = 20+ 2.03 = 22.03 oC

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