20.0 mL of ethanol initially at 7.2 degrees Celsius is mixed with 31.7 mL of water initially at 28.9 degrees Celsius in an insulated beaker. Assuming no heat is lost, what is the final temp?

I did not include the density, but I solved for mass, and have the SHC, for both. Don't know what to do from there, since I don't have q

Assuming no heat of solution...

Heat gained by ethanol+heatgaiedwater=0

assuming density of each is same..
20*Cthan(tf-7.2)+31.7(cwater)(tf-28.8)=0
look up the specific heats of ethanol, and water, but assuming they are the same, you have and easy calculation to find final temperature tf

I do not know how to set that up

Jennifer, Bob P set it up for you. All you need to do is to solve for Tf (and look up specific heats).

How do I solve for Tf?

I'm so confused

20.0 ml of ethanol (d=0.789 g/ml) initially at 7.2c is mixed with 31.7 ml of water (d=1.0g/ml) initially at 28.9c in an insulated beaker. Assuming that no heat is lost, what is the final temp of the mixture?

I've got mass of both and specific heat capacity
Do I put 0 for q?
And I don't know how to solve for Tf.

Bob P set up the chemistry part. What you are having trouble with is the math.

No, you don't put zero for q. You don't need to solve for q. The heat lost by one material + the heat gained by the other material must = to zero and that's the way Bob P set up the problem. Look at the first part.
20*Cthan(tf-7.2). Let's say Cethanol is about 4. So now
20*4(tf-7.2). You solve this by clearing the parentheses.
80(tf-7.2) = 80tf - 80*7.2 = 80tf-576. Do the second part the same way, combine like terms, and solve for tf. You shouldn't be confused with tf. That's X in the equation, the unknown. You want to know the final T. You do the second part and combine terms etc and I'll check your work.