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

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Homework Help Overview

The discussion revolves around a calorimetry problem involving a heated copper sample and water, focusing on the calculation of the mass of water based on heat transfer principles. The original poster presents a scenario where a 110-g sample of copper at a specific temperature is placed in water at a different temperature, aiming to find the mass of the water when thermal equilibrium is reached.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the principle of conservation of energy using the formula for heat transfer, but expresses confusion regarding their calculations. Some participants question the accuracy of the numerical calculations and the treatment of significant figures.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the problem, including potential errors in calculations and the application of significant figures. There is no clear consensus on the correct mass of water, as participants are still analyzing the problem.

Contextual Notes

Participants mention issues related to significant digits and the accuracy of numerical results, indicating that the problem may involve constraints related to precision in measurements.

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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