How Does Temperature Equilibrate in Mixed Metal and Water Systems?

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Homework Help Overview

The discussion revolves around a thermal equilibrium problem involving an aluminum vessel, water, and a piece of iron, focusing on how temperature equilibrates in mixed metal and water systems.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different approaches to calculating the final temperature and question the specific heat values used in their calculations. Some participants express uncertainty about the units and meaning of the equations being applied.

Discussion Status

Multiple interpretations of the specific heat values and their impact on the final temperature are being explored. Some participants have provided calculations that yield different results, and there is an ongoing examination of the equations and their units.

Contextual Notes

There is mention of potential discrepancies in specific heat values and the need to reference textbook examples for clarification. Participants are also discussing the implications of using different units in their calculations.

Karol
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Homework Statement


An aluminum vessel of 500 gr contains 117.5 gr water at 200C. a piece of iron of mass 200 gr and 750C is thrown inside. what's the final temperature and the water equivalent of the vessel.

Homework Equations


Specific heat of aluminum: 0.217
Specific heat of iron: 0.113

The Attempt at a Solution


##(500\cdot 0.217+117.5)(t-20)=200\cdot 0.113 (75-t)\rightarrow t=25^0##
The answer should be 230C.
The water equivalent:
##500\cdot 0.217=m=108.5 gr##
The answer should be 110 gr
 
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I make it 25°. The difference may be explained by use of S.H. values slightly different from those used by the textbook authors?
 
I'm seeing slightly different values for relative specific heats depending on where people round off or what standard they use or something or other.
I'm also thinking - check the book values from the chapter or any examples they give.
 
For the aluminum container... 500 g * 0,217 * 293 K = 31791
For the water...... 117,5 g * 1 * 293 K = 34428
For the iron...... 200 g * 0,113 * 348 K = 7865

Total ............. = 74084

Now, 74084 / ((500 *0,217)+(117*1)+(200*0,113)) = 298,6 K = 25,6 ºC
 
I don't know this equation: mass (x) specific heat (x) deg. kelvin, what are the units of the result? calories? or is it just a mathematical trick, some kind of a mean.
I only know: mass (x) specific heat (x) \Delta t
 
Karol said:
I don't know this equation: mass (x) specific heat (x) deg. kelvin, what are the units of the result? calories? or is it just a mathematical trick, some kind of a mean.
I only know: mass (x) specific heat (x) \Delta t

It's a 'weighted mean'. Useful for a lot of things. The units don't matter, but they are cal, since [grams * (cal/(grams * K)) *K] simplify to cal...
 
Thanks
 

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