Thermodynamics question: Please help with number 17 in the pdf attachment. Its like their skipping a step on how to get final temperature.

Question

happykiddo · Answer

http://www.elcamino.edu/faculty/pdoucette/Thermochemistry-practice-problems-key.pdf

frostbite · Answer

No not really. They simply isolate $T_f$ in the equation as all variables except Tf are known.

frostbite · Answer

If you can explain what kind of step you feel you are missing. I'll might be able to provide a better explanation.

happykiddo · Answer

I get up to this...
250 J= 1.47(Tf-24 deg C)
I distribute the 1.47 and solve for Tf
But I dont get the 28.3 deg C they do

happykiddo · Answer

Can you help me understand?

frostbite · Answer

I sure will. Let me fast get a pen and paper :)

frostbite · Answer

Hmmm yeah I can see the issue.
$$\Large q=m c \Delta T $$$$\Large q=mc(T_2-T_1)$$$$\Large T_2=T_1+\frac{ q }{ mc }$$
But all the units cancel with the information so all the units are correct. So it seems like your teacher gave you a bad key.

frostbite · Answer

I can't even figure out what error he may have done that gives a $\Delta T=4.3^{\circ} \sf C$

frostbite · Answer

But I think your teacher is incorrect in his answer. Might be a typo or calculation error, no matter what the information and the result does not agree with each other :)

happykiddo · Answer

Thank you, I started freaking out because I test on this tuesday. Really appreciate your time. I'll ask my professor.