1. The problem statement, all variables and given/known data A 155-g piece of copper at 188oC is dropped into 250.0 g of water at 23.7oC. (The specific heat of copper is 0.385 J/goC.) Calculate the final temperature of the mixture. (Assume no heat loss to the surroundings.) Knowns. 155g Copper Initial Temp 188 C 250.0g Water Initial Temp 23.7 C Spec Heat Copper .385 J/G C Spec Heat Water 4.184 J / G C Unknown Final Temp Water C - Celsius J - Joules G - Grams T - Final Temp 2. Relevant equations Heat lost by copper = Heat gained by Water my equation setup 155g × 0.385J / g °C × (Tfinal-188 °C) = 250.0g × 4.184j / g °C × (Tfinal-23.7 °C) 3. The attempt at a solution Well I have several pages of attempts here but I'll take a screenshot of my most recent. Oddly enough I took a guess on the system we use (wileyplus) to get a hint as it gives you a hint on attempt 2 and got it correct. But it's bugging me that I can't solve it even after repeated attempts. imageshack.us/photo/my-images/38/hwquestion.jpg/ ^ wiley Plus screen saying my 32.5 guess was correct / This attempt I dropped the units and just solved for T Someone in the class mentioned trying this method but the result is off or I'm completely brain dead this evening and failing at math / 155g x 0.385 J/g C x (Tfinal-188 C) = 250.0g x 4.184 J/g C x (Tfinal-23.7 C) 59.675 x (t - 188) = 1046 x (t - 23.7) 59.675t - 11218.9 = 1046t - 24790.2 -986.325t = -13571.3 t = 13.759 Which would be way to far off from the +-1 sigfig on the 3rd digit allowance that said 32.5 C was correct. When I go for the dimensional analysis route I can't isolate Celsius, I always end up with joules as a unit on my final answer. 155g | .385 J | (T - 188 C) | = 250g | 4.184 J | (T - 23.7 C) | g C | | | g C | Gives me the same result with Joules left. I'm totally lost here, the bad part of my brain keeps saying you guessed the right answer so move on, but I can't leave this alone I have to know how to solve it. Any help would be much appreciated.