posted by james on .
An unknown volume of water at 18.2°C is added to 27.0 mL of water at 36.7°C. If the final temperature is 23.5°C, what was the unknown volume? (Assume that no heat is lost to the surroundings; d of water is 1.00 g/mL.)
heat lost by one + heat gained by the other = 0
(mass1 x specific heat x [Tfinal - Tinitil]) + (mass2 x specific heat x [Tfinal - Tinitial]) = 0
You have only one unknown, solve for mass (1 or 2 depending upon how you set up the problem), then since the density of water is assumed to be 1.00 g/mL, the mass will be the same as the volume, in mL. Post your work if you get stuck.
please explain more