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Calorimetry

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By Expert Faculty of Sri Chaitanya
Question

M gram of steam at 100°C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40ºC [heat of vaporization of water is 540 cal/g and heat and heat of fusion of ice is 80 cal/ g], the value of M is_______

Moderate
Solution

Heat gain by ice = Heat loss by steamMiceLf+Mice400Cw=msteamLV+msteam10040Cw20080+401=M540+601200120=M600                    M=40g


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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 S=1+θ500 cal g1C1where θ is temperature in oC. The specific heat capacity of ice, water and the calorimeter remains constant and values are Sice =0.5 cal g1C1Swater =1.0 cal g1C1 and latent heat of fusion of ice is Lf=80calg1. Assume no heat loss to the surrounding and calculate the value of m in grams. 

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