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## hot metal + water = Tf???

Yet again I find i am stumped by having no clue as to what I did in this incredibly easy problem.

 A 110.-g sample of copper (specific heat capacity = 0.20 J/ºC•g) is heated to 82.4 ºC and then placed in a container of water at 22.3 ºC. The final temperature of the water and copper is 25.6 ºC. What is the mass of the water in the container, assuming that all heat lost by the copper is gained by the water?
Simple right?

1 = water
2 = copper

m1c1 delta T1 = -m2c2 delta T2

This simplifies to...

m1= (-m2c2 delta T2)/(c1 delta T1)

which is...

(-(110*.2*(25.6-82.4)))/((4.18*(25.6-22.3)) = 95.8533g

Computer says i'm wrong and it slapped me

I then did the lazy way and calculated the energy required to lower the copper's temperature. With that, I manually figured out how much water there was and got the exact same answer.

Where am I going wrong here?
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 Significant digits? I don't know...
 Recognitions: Gold Member Homework Help Did you do your numerical calculation right? Isn't the answer 90.59 g?

Recognitions:
Gold Member

## hot metal + water = Tf???

I tried 3 and 4 sig. figures with +/- 1 each direction and nothings coming out correctly. God i wanna pound the writers of this textbook.

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