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 $$S=1+$$θ$$500$$ cal g−$$1$$∘C−$$1where$$ θ is temperature in oC. The specific heat capacity of ice, water and the calorimeter remains constant and values are Sice $$=0.5$$ cal g−$$1$$∘C−$$1$$; Swater $$=1.0$$ cal g−$$1$$∘C−$$1$$ and latent heat of fusion of ice is $$Lf=80calg$$−$$1$$. Assume no heat loss to the surrounding and calculate the value of m in grams.

Question

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 $$S=1+$$θ$$500$$ cal g−$$1$$∘C−$$1where$$ θ is temperature in oC. The specific heat capacity of ice, water and the calorimeter remains constant and values are Sice $$=0.5$$ cal g−$$1$$∘C−$$1$$; Swater $$=1.0$$ cal g−$$1$$∘C−$$1$$ and latent heat of fusion of ice is $$Lf=80calg$$−$$1$$. Assume no heat loss to the surrounding and calculate the value of m in grams.

Infinity Learn · Accepted Answer

Heat gained by ice    $$=mSice$$ ×$$10+mSm$$×$$20$$×$$mLf=105m$$.calHeat lost by calorimeter $$=10$$×$$1$$×$$20=200calHeat$$ lost by $$liquid=$$−$$50$$∫$$4020$$ $$1+$$θ$$500d$$θ$$=50$$θ$$+$$θ$$210002040=5040+16001000$$−$$20+4001000=50$$×$$21.2=1060cal$$∴$$105m=1060+200$$⇒$$m=12g$$

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