# Heat exchange in a bath

1. Oct 15, 2014

### Karol

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Specific heat of aluminum: 0.217
Specific heat of iron: 0.113

3. The attempt at a solution
$(500\cdot 0.217+117.5)(t-20)=200\cdot 0.113 (75-t)\rightarrow t=25^0$
The water equivalent:
$500\cdot 0.217=m=108.5 gr$
The answer should be 110 gr

2. Oct 15, 2014

### Staff: Mentor

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

3. Oct 15, 2014

### 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.

4. Oct 15, 2014

### 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

5. Oct 15, 2014

### 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) $\Delta t$

6. Oct 15, 2014

### NTW

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...

7. Oct 15, 2014

Thanks