80 gm of water at 30oC is poured on a large block of ice at 0oC. The mass of ice that melts is

Since, the block of ice at 0oC is large, the whole of ice will not melt, hence final temperature is 0oC .∴Q1 = heat given by water in cooling upto 0oC=msΔθ=80×1×(30−0)= 2400 calIf m gm be the mass of ice melted, thenQ2=mLF=m×80  Now, Q2=Q1m x 80 = 2400 or m = 30 gm.