A piece of ice with a mass of 60 g is transferred directly from the freezer at a temperature of –7°C to a polystyrene cup containing 200 g of water at 75°C. Calculate the final temperature of the water. Ignore any heat loss to the cup and its surroundings. Qheat ice + Qmelt ice + Qheat cold water + Qcool hot water mciceΔtice + mLf + mcwΔt + mhwcwΔthw = 0 (0.060 kg) x (2100 J/(kg.C)) x (0°C - (-7°C)) + ( 0.060 kg) x (3.3 x 10^5) + (0.060 kg) x (4200 J /(kg.C)) x (t2 – 0°C) + (0.2) x (4200 J) x (t2 – 75°C) = 0 882 + 19800 + 252t2 + 840t2 – 63000 = 0 840t2 = 42318 t2 = 50°C The above problem is from my book, but I am having difficulty with the equation. Where did the 63000 come from?