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March 29, 2015

March 29, 2015

Posted by **Mandi** on Friday, April 27, 2012 at 4:18am.

(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?

- Physics -
**Elena**, Friday, April 27, 2012 at 7:55am1J =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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