What Is the Correct Mass of Water When Hot Copper Meets Cold Water?

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SUMMARY

The discussion revolves around calculating the mass of water when a heated copper sample is placed in it. A 110 g copper sample with a specific heat capacity of 0.20 J/ºC•g is heated to 82.4 ºC and placed in water at 22.3 ºC, resulting in a final temperature of 25.6 ºC. The correct mass of water calculated using the formula m1 = (-m2c2 delta T2)/(c1 delta T1) yields approximately 95.85 g, but the user questions the accuracy of this result, suspecting issues with significant figures or numerical calculations. The user also mentions that an alternative method yielded the same mass, indicating a potential misunderstanding of the calculation process.

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Pengwuino
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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?
 
Last edited:
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Significant digits? I don't know... :confused:
 
Did you do your numerical calculation right? Isn't the answer 90.59 g?
 
I tried 3 and 4 sig. figures with +/- 1 each direction and nothings coming out correctly. God i want to pound the writers of this textbook.
 

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