Part A A large punch bowl holds 3.60 kg of lemonade (which is essentially water) at What is the final temperature of the system? Ignore any heat exchange with the bowl or the surroundings. 24.0 . C. A 5.80x10-2-kg ice cube at -10.0 . C is placed in the lemonade. View Available Hint(s) T = 22.3 .C Submit Previous Answers Correct Picture the Problem: Ice cubes are placed in a bowl of lemonade. Heat transfers from the lemonade to the ice, causing the ice to melt until the ice and lemonade arrive at the same temperature. Strategy: Assume that all of the ice melts and that the lemonade and melted ice arrive at an equilibrium temperature between 0 . C and 24.0 . C. The heat absorbed by the ice will raise its temperature to 0 C, melt it, and then raise the temperature of the melted water to the equilibrium temperature. Set the amount of heat absorbed by the ice equal to the amount of heat given off by the lemonade as it cools to the same final temperature. Solve the resulting equation for the final temperature. Solution: Use equations c = MAT and L = to calculate the heat gained by the ice: Qi = Cice mice Tice + mice Lf + miceCw(Ti -0 .C) = (5.80 x 10-2 kg) ([2090 J/(kg . K)](10.0 .C) + 3.35 x 105 J/kg) + miceCw If Qi = 2.06 x 10* J + mice Cw If. Calculate heat lost by the lemonade: Qw = Mlem Cw (24.0 . C - If). Set the two heats equal: Qi = Qw = 2.06 x 10* J + miceCwTi = MlemCw (24.0 .C - TF ). Solve for the final temperature: T=" mlem Cw (24.0 .C)-2.06x104 J (3.60 kg) [4190 J/(kg.K)](24.0 .C)-2.06x104 J mice +mw ) Cw = 22.3 . C. 5.80x10-2 kg+3.60 kg) [4190 J/(kg.K)] V Part B What is the amount of ice (if any) remaining? Express your answer using one significant figure. View Available Hint(s) JA AEd m = kg