A 225g sample of water at 35.4ºC is added to 334g of water at 20.7ºC in a thermally insulated container.

what will be the final water temperature?
a.28.0ºC
b.9.64ºC
c.26.6ºC
d.29.5ºC
e.-8.9ºC

please write the procedure

When volumes of the same material (same specific heat capacity) are mixed together, the final equilibrium temp attained becomes quite simple to calculate. Multiply the mass of each times its temperature and then add them together. Then divide by the total mass of the mixture.(mc= mass of cold, mh = mass of hot, Tc = cold temp, Th = hot temp , Tf - final temp)
Tf = [(mc * Tc) + (mh * Th)] / mc + mh
= (334-g x 20.7 oC) + (225-g x 35.4 oC) / 559-g
Tf = 26.6 oC

You have all the information you need to solve this question.

The general set up is:

mc(Tf - Ti) = -(mc(Tf - Ti))

Where:

m = mass of the water (left hand mass is 225g and and the right hand mass is 334)
c = Specific heat of water (this will cancel out because it is the same on both sides of the equation).
Tf = final temperature (what we are looking for)
Ti = Inital temperature (This is 35.4C on the left and 20.7C on the right)

Now just plug in the numbers and solve for Tf

225g(Tf - 35.4C) = -(334g(Tf - 20.7C))

Do the algebra and you'll end up with

Tf = 26.6 C

The answer is C

can you add anymore information about the question and I may be able to help