Questions
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
detailed solution
Correct option is B
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=503gTalk to our academic expert!
Similar Questions
A calorimeter of water equivalent 10 g contains a liquid of mass 50 g at 40oC. When m gram of ice at -10oC is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be 20oC. It is known that specific heat capacity of the liquid changes with temperature as where is temperature in oC. The specific heat capacity of ice, water and the calorimeter remains constant and values are ; and latent heat of fusion of ice is . Assume no heat loss to the surrounding and calculate the value of m in grams.
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