This is a pretty straightforward heat/energy exchange problem, but I can't figure the answer out.

A constant-pressure calorimeter contains $250.0\ \mathrm{mL}$ of ethanol (density = $7.89\ \mathrm{g\ mL^{-1}}$ and specific heat capacity = $2.460\ \mathrm{J\ g^{-1}\ ºC^{-1}}$ at $22.5\mathrm{ºC}$. If a $44.7\ \mathrm{g}$ sample of tin (specific heat capacity = $0.210\ \mathrm{J\ g^{-1}\ ºC^{-1}}$ at $172.3\mathrm{ºC}$ is dropped into the calorimeter, then what will be the highest temperature reached by the ethanol?

So I worked this problem using $q = m c \Delta T$, setting $q$'s equal between both sides since the heat energy leaving the tin will be absorbed by the water... but I just now realized as I was typing this, that only the $\Delta q$ is the same for both sides, not the total $q$. I tried working this problem a couple times but kept getting $22.2\mathrm{ºC}$ for both substances, which doesn't make sense since the final temperature should be above $22.5\mathrm{ºC}$. my attempt at solving this thing

EDIT: I added the above photo... I took the advice of one of the repliers (sorry, I would give you credit but I can't see your name while posting this). I had both sides as positive, whereas one should be negative, as you can see in my picture. Using this method, I got 25.344 degrees C, which is a very reasonable answer since the ethanol had the higher specific heat and greater mass... we would expect to see a number much closer to 22.5 than 172.3.

I believe this is the corrent answer. If anyone wants to doublecheck my figures, please go ahead and LET ME KNOW if you find anything wrong. Thanks!

  • $\begingroup$ Welcome to Chem.SE! You should show us one of your attempts so we can better assist you where you went wrong. Otherwise, there's no way for us to help you, really. $\endgroup$ – ringo Dec 3 '16 at 5:54

Density wrong for ethanol. Fix it. Fix the mass.

$$Δq = m c ΔT$$

ΔT of ethanol is $T_{final_{system}}$ -$T_{initial_{ethanol}}$


ΔT of tin is $T_{initial_{tin}}$-$T_{final_{system}}$

Notice that $T_{final_{system}}$ and -$T_{final_{system}}$ are on opposite sides of the full equation.

You will get $T_{final_{system}}$ by itself.

Use the distributive property

If you want to show your $ΔT$s and $Δq$s as being only the magnitude of the change, and not the sign (ie no negative $ΔT$s or $Δq$s) then you should use $T_{initial_{tin}}$-$T_{final_{system}}$ on the right side and remove the - from the right side. (Though you could then pull -1 out and put it into your terms).

Also, you should limit your final answer to 3 significant figures.

Lastly, if it is important to show your work, you might want to include all of your units in line 4 and your units for temperature throughout the process.

  • $\begingroup$ Please don’t use answers full of ‘hint:’ on chemistry. Either answer or not. If you need a spoiler tag, add it using >! (mind the space) at the beginning of a line. $\endgroup$ – Jan Dec 3 '16 at 12:55
  • $\begingroup$ Changed. Note that prior to his recent edit, the question was about what steps he needed to follow to proceed. $\endgroup$ – Joseph Hirsch Dec 3 '16 at 15:02

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