Specific heat/equilibrium in three component system (Thermo)

[RadiumBlue]

Homework Statement

Rather than retyping the problem, I've uploaded a screenshot here >> http://imgur.com/EB5MrtP

Homework Equations

$Q_1 + Q_2 + Q_3 ... = 0$
$mc \Delta T = Q$

The Attempt at a Solution

a. $Q_{tea} + Q_{met} + Q_{cr} = 0$

I converted the temperatures to K, so $T_{cr}$ and $T_{met}$ = 293.15K, while $T_{tea}$ = 373.15K

Constructing the equation, I got:

$0 = (m_{met} * c_{met}* (T_{equil}-293.15)) + (m_{cr} * c_{cr} * (T_{equil}-293.15)) + (m_{tea} * c_{tea} * (T_{equil} - 373.15))$

I'm stuck here now, because the problem doesn't give any of the masses, it only relates them to each other. The specific heats I can find because the specific heat of water is a known value, but I'm stuck on the masses. What am I missing?

 

[TSny]

I believe the problem is asking you to express the equilibrium temperature in terms of the symbols for the masses and specific heats. If so, you do not need to plug in numerical values for these quantities.

 

[RadiumBlue]

I believe the problem is asking you to express the equilibrium temperature in terms of the symbols for the masses and specific heats. If so, you do not need to plug in numerical values for these quantities.

Oh, there's a small part at the bottom that got cropped out, it specifically asks for a numerical value. It also mentions something about using a binomial expansion to solve it, which I didn't understand how it applied to this problem. I've amended the link in the OP with the fixed version.

 

[TSny]

OK. Sorry, I did not see part (c) when I first looked at the problem.

Try expressing all masses in terms of one of the masses, say $m_{met}$. Likewise, express all specific heats in terms of one of the specific heats, say $c_{met}$.

You will then find that the equation simplifies such that you do not need any numerical values for the masses or specific heats.

Also, you do not need to express the temperatures in K. You can keep them in ^oC because you are dealing with temperature differences here.

 

[RadiumBlue]

Oh, duh, I see it now. Thank you! I probably should have tried that first, but I just didn't think of it.