assume the final temp is 0C (we can check that later)
heat to melt ice+Heat to freeze water=0
68*specific heat of ice*(0- (-23)+Mass*Hf=0
mass= 68*.5*23/80=9.78 grams
check my thinking.
A 68 g ice cube at −23◦C is dropped into a container of water at 0◦ C.
How much water freezes onto the ice? The specific heat of ice is 0.5 cal/g ·◦ C and its heat of fusion of is 80 cal/g.
Answer in units of g.
1 answer
1 answer