Please Help! - Specific Heat Problem... I have been stuck for a while, it's online (Webassign), so it told me my answer was wrong. ------- A 6.3x10^2 g sample of water at 90.0°C is mixed with 4.45x10^2 g of water at 22.0°C. Assume no heat loss to the surroundings. What is the final temperature of the mixture? ------- The formula I know is right: (cp)(m)(Tf- Ti) = (cp)(m)(Tf- Ti) cp= spec. heat m = mass Tf = final Temp Ti = initial Temp -------- work: converted grams into kilograms... cp water, given || (4186) (.630) (x-90) = (4186) (.445) (x-22) 774.41x = 196365.26 x = <<<it says it's wrong>>> Where did I slip up? Thanks.