What is the final temperature of a mixture of 250 g of water at 33°C into which 3 g of ice at -20°C is placed
First things first. Look up heat fusion for ice, specific heat ice and specific heat wq to change T of ice from -20 to 0 C is
q1 = 3 g x 2.08 J/g*c x [(0 - (-20)] = estimated 125 J.
q2 = to change solid ice to liquid water is
q2 = 3 g x 334 J/g = estimated 1004 J
q3 = heat added to change T from -20 to 0 + heat to melt ice @ 0 = 125 + 1002 = 1127 J so we have 3 g ice now at 0 C mixed with the 250 g H2O @ 33 C. Put these together.
250 g x 4.184 J/g*c x (Tfinal - 33) + [3g H2O x 4.184 J/g*c x (Tfinal-0)] + 1127 = 0
Solve for Tfinal. Post your work if you get stuck. ater. Then .......