If 8162.7 J of heat are added to 39 g of water initially at 19°C, (a)...

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Question

If 8162.7 J of heat are added to 39 g of water initially at 19°C, (a)...

If 8162.7 J of heat are added to 39 g of water initially at 19°C,
(a) How much energy is this in calories?
(b) What is the final temperature of the water?

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Answer 1

Answer #1

a)

1 cal = 4.18 J

1 J = 1/ 4.18 cal

8162.7 J = 1952.8 cal

=====

b)

q = m C dT

8162. 7 = 39*10^-3* 4186* dT

T - 19 = 50

T = 69 C

======

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Answer 2

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