Recent content by prwraith

Just to clarify 155g × 0.385J / g °C × (188°C - Tfinal) = 250.0g × 4.184j / g °C × (Tfinal-23.7 °C) The underlined portion is now correct because we're looking for the amount of heat lost en route from the original temperature to the final temperature. And the bold portion is correct because...

Well after you point it out it does actually make sense haha. Why in gods name would I be subtracting my highest temperature, something that's clearly going to go down after being plunged into water from the final temperature which I already knew was far far lower. This half of the equation...

Yay I got the answer ! 155g x .385 J / g C x (188C - T) = 250g x 4.184 j / g C x (T - 23.7C) -59.675 T + 11218.9 = 1046 T - 24790.2 -1105.675 T = -36009.1 C = 32.5 Yay I love you for the help with switching that sign on my equation !

hmm ok let me run the numbers, I just ran the numbers on switching the sign to + while waiting for a reply and got 36.5 which is wayyy closer. Ty for the help, let's see what this round gives.

by wrong sign do you mean the -?

Homework Statement 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...