Question

a punch bowl holds 3.99kg of lemonade (which is essentially water) at 20.5 C. A 0.0550kg ice cube at -10.2 C is placed in the lemonade. What is the final temperature of the system, and the amount of ice (if any) remaining? Ignore any heat exchange with the bowl or the surroundings

Answer 1

find the amount of heat it takes to warm .0550 kg of ice from -10.2 to 0 deg.
find the amount of heat it takes to melt .0550 kg of ice (use heat of fusion)
add them and use in
-delta Q = mc(delta T) (ie remove the heat it takes to warm and melt the ice from the lemonade, m is the mass of the lemonade, c is the specific heat of lemonade and delta T is Tf -20.5)

then you're left with .0550kg of water at 0 degrees and the lemonade at whatever final temp you found above
then calculate the equilibrium temp (T) that they come to:
.0550kg(T) = 3.99kg( Tf-T)