Question

A calorimeter contains 400g of water at a temperature of 5∘C. Then, 200g of water at a temperature of +10∘C and 400g of ice at a temperature of −60∘C are added. What is the final temperature(in oC) of the contents of the calorimeter? Specific heat capacity of water =1000cal/kg/K. Specific latent heat of fusion of ice =80×1000cal/kg. Relative specific heat of ice =0.5

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Solution

Let t be the common temperature above zero celsius. Heat lost by calorimeter and water added=400×10−3×1000(5−t)+200×10−3×1000(10−t)Heat gained by ice=400×10−3×5000(60+t)+400×10−3×80000+400×10−3×1000tSince Heatlost=heatgained∴400(5−t)+200(10−t)=200(60+t)+400×80×+400tor t=−24.33∘CThis is contradictory to the assumption that t is above zero. Hence, the final temperature is either 0∘C or less than 0∘C. Let the temperature be t∘C (less than 0∘C).Then 400×10−3×1000×5+200×10−3×1000×10+600×10−3×80000+600×10−3×500×t=400×500(60−t)or 400×5+200×10+600×80+300t=200(60−t)or 20+20+480+3t=120−2tor t=−80∘CThis is absurd because the lowest temperature is −60∘C and we discard this second assumption also.So the final temperature is 0∘C.Let x grams of water be converted into ice.Heatlost=400×10−3×100×5+200×10−3×1000×10+x×10−3×8000=4000+80xHeatgained=400×10−3×500×60or 4000+80x=12000or x=100gHence, the final result is 500g of ice and 500g of water at 0∘C.

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