How Does Temperature Equilibrate in Mixed Metal and Water Systems?

[Karol]

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

 

[NascentOxygen]

I make it 25°. The difference may be explained by use of S.H. values slightly different from those used by the textbook authors?

 

[Simon Bridge]

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.

 

[NTW]

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

 

[Karol]

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) $\Delta t$

 

[NTW]

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...

 

[Karol]

Thanks