[trogdor3491]

 A silver spoon with a mass of 99g and temperature of 280K (Silver's specific heat is .235 J/(g*C)) is placed into a Styrofoam cup containing 200mL (200g) of coffee at 350 K (assume same specific heat 4.18 J/(g*K))

What is the final temperature of the coffee?


 I think I figured out a way to do it but I can't complete it:
q= energy lost/gained
cp= specific heat
m= mass
^T = difference in temperature difference between initial and final
q= cp * m * ^T

The energy lost and gained should be equal so you can set this right? (I'm pretty sure we assume no heat lost to surroundings.)
.235*99*(final temp-280) = 4.18 * 200*(350-final temp)

I'm missing the final temp though so thats a pretty big problem, anyone got ideas?

[enahs]

First, remember the change in temperature is always (Final - Initial), you have it reversed in one case.


But no, missing the final temp is not a big problem. This is what you want to know, you know everything else! One equation and one unknown! You just have to use algebra and solve for the final temperature.

[trogdor3491]

I took:
.235*99*(final temp-280) = 4.18 * 200*(350-final temp)

and made:
.235*99*(x-280) = 4.18 * 200*(350-x)

I think deltaT needs to be positive because when I solve it like this:
1. multiply the numbers outside the parenthesis= 23.265(X-280) = 836(350-X)
2. Distribute throughout- 23.265X - 6514.2 = 292600 - 836X
3. Get Xs on one side- 859.265X = 299114.2
4. Solve for X- X= 348.1047
I get an answer that makes sense, but if I just made it always final - initial then I get 352K which doesn't make any sense.

And when I use q= cp*m*deltaT for the silver and coffee they the amount of joules lost/gained is the same.

So is it ok that I made deltaT = initial - final?

[Sev]

Yes, that's what I would do 

[enahs]

So is it ok that I made deltaT = initial - final?

No.
ΔT is always Final-Initial.

While it makes it mathematically correct when you solve for x, it is more important to understand what is going on here.

The key here is that there is the assumption that no heat is lost, that is.
Q₁+Q₂=0, or Q₁=-Q₂
Any heat lost by one is gained by the other, or vice-versa.

With the appropriate negative sign, the Final-Initial works; and how it should be worked.

[Sev]

With -ve it is same method: -[mC(T-350)]=mC(350-T)

[enahs]

With -ve it is same method: -[mC(T-350)]=mC(350-T)

Mathematically it is the same. Conceptually it is different.

Say we have a system of 4 parts:
A calorimeter, Water, Ice and a piece of metal.
Do you really want to fuss around with making some Final-Initial and others Initial-Final?

Or use:
Q1+Q2+Q3+Q4=0 and use Final-Initial is much easier and reliable.