Heat Capacity iron heating water

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Discussion Overview

The discussion revolves around calculating the experimental heat capacity of iron based on a specific experiment involving heating water. Participants explore the application of relevant equations and the methodology for solving the problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • Bjorn presents the experimental setup and asks for help in calculating the heat capacity of iron.
  • One participant prompts Bjorn to look up the relevant equation in his textbook, emphasizing the importance of attempting the problem independently.
  • Bjorn shares his calculations, outlining the mass of iron and water, initial and final temperatures, and the specific heat capacity of water, leading to a proposed value for the heat capacity of iron.
  • Another participant affirms Bjorn's calculations as correct and encourages him for solving the problem on his own.

Areas of Agreement / Disagreement

There is no explicit disagreement among participants, but the discussion reflects varying levels of engagement with the problem-solving process. Bjorn's calculations are acknowledged as valid by another participant.

Contextual Notes

Participants do not address any assumptions or limitations in the calculations, nor do they clarify the definitions or conditions under which the heat capacity is being calculated.

Who May Find This Useful

This discussion may be useful for students learning about heat transfer, specifically in the context of calculating heat capacities in experimental settings.

bjorn
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This is what I've got:

Iron mass: 81.9g
Initial iron temp.: 92'C
Water volume: 100mL
Initial water temp.: 25'C
Final water temp.: 29'C


What the the experimental heat capacity of iron?


Thanks :)

Bjorn
 
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Bjorn, do you know what equation is to be used for this kind of problem. It's very simple and will be in your textbook. Look it up.

We can not do your homework for you. You have to make an attempt first.
 
Hmmm...

How about this, does this seem correct?


mm = 81.9g

mw = 100mL

Tim = 92°C

Tf = 29°C

Tiw = 25°C

Tfw = 29°C

cw = 4.19 J/g·°C (this is the number the school operates with)

cm = ?


81.9g(cm)(29°C - 92°C)+100g(4.19J/g·°C)(29°C - 25°C) = 0

81.9g(cm)(-63°C)+100g(4.19J/g·°C)(4°C) = 0

-5159.7g·°C(cm)+1676J = 0

1676J = 5159.7g·°C(cm)

Divide both sides by 5159.7g·°C:

cm = 0.325J/g·°C


Didn't take this from the textbook, I guess I knew what to do, but I have always been an idiot with unkowns in math :smile:
Equations and me aren't exactly best friends.
 
This is perfectly good.

Aren't you glad you did it by yourself, rather than have someone else do it for you ?
 

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