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

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