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Energy in the amount of 420 J is added to a 35 g sample of water at a temperature of 10.0 C What is the final temperature of the water?
Wiki User
∙ 13y ago
For energy problems like this the following equation is used:
E = m*c*(T2-T1) where
E is the energy amount added
m is the mass of the thing being heated
c is the specific heat of the object being heated
T2 is the temperature after the heat
T1 is the temperature before the heat is added.
If you're not familiar c, the specific heat of a substance is a property which is a measure of how much heat is required to heat 1 gram of the substance by 1 degree Celsius. For water it is 1.0 cal/g/C. Since there are 4.186J per calorie you could also use c=4.186 J/g/C.
You can rearrange the above equation to solve for T2:
T2 = T1 + E/(m*c) = 10C + (420J)/((35g)*(4.186J/g/C)) = 13 C
I hope this helps
Wiki User
∙ 13y ago
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