10 g of ice at -20oC is dropped into a calorimeter containing 10 g of water at 100C; the specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain

Q1=10×1×10=100calQ2=10×0.50[0−(−20)]+10×80     =(100+800)cal=900caltemperature = 0oCAs heat given by water in cooling up to 0oc is only just sufficient to increase the temperature of the ice from - 20oC to 0oC, hence mixture in equilibrium will consist of 10 g of ice and 10 g of water, both at 0oC.