10 gm of ice at -20oC is dropped into a calorimeter containing 10 gm of water at 10oC; the specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain

Q1=10×1×10=100calQ2=10×0.5[0−(−20)]+10×80=(100+800)cal=900calAs Q1 < Q2, so ice will not completely melt and final temperature = 0oC.As heat given by water in cooling upto 0oC is only just sufficient to increase the temperature of ice from -20oC to 0oC hence mixture in equilibrium will consist of 10 gm of ice and 10 gm of water, both at 0oC.