Ice at 0oC is added to 200 g of water initially at 70oC in a vacuum flask. when 50 g of ice has been added and has all melted, the temperature of flask and contents is 40oC. When a further 80 g of ice is added and has all melted, the temperature of whole becomes 10oC. Neglecting heat lost to surroundings the latent heat of fusion of ice is

According to principle of calorimetry,MLF+MsΔT=(msΔT)water +(msΔT)flask 50LF+50×1×(40−0) =200×1×(70−40)+W(70−40)50LF+2000=(200+W)305LF=400+3W                   .........(i)Now the system contains (200 + 50) g of water at 40oC, so when further 80 g of ice is added80LF+80×1×(10−0)       =250×1×(40−10)+W(40−10)80LF=670+3W    .........(ii)Solving Eqs. (i) and (ii),LF=90cal/g  and  W=503g