Asked by Summer

Use the heating curve of 50 g of water and the list of values to answer the question.

specific heat of ice = 2.10 J/(g·°C)
specific heat of water = 4.18 J/(g·°C)
specific heat of water vapor = 2.07 J/(g·°C)
latent heat of fusion of ice = 333.4 J/g
latent heat of vaporization of water = 2256 J/g

If 9200 J of heat were added to the water as depicted by the heating curve, what is the final temperature of water?
Answered by DrBob222
I don't know what the heating curve looks like and I don't know what that has to do with the problem.
Answered by Madison
I tried 34 C and I got it wrong...
Answered by Storm
Multiply the specific heat of water by the grams of water you are given.
(4.18J/g*C)(50g)=209J/C
Then, divide 9200J by 209J/C to get 44C
Answered by so....help_me
it wasn't 44 either
Answered by ;-;
its 54 C
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