14A block of ice at0°C is added to a 150. g aluminum calorimeter cup that holds2

Question

14A block of ice at0°C is added to a 150. g aluminum calorimeter cup that holds200. g of water at 10°C. If all but 2. g of ice melt, what wasthe original mass of the block of ice?   

A) 38.8 g B) 62.2 g C) 47.6 g D) 31.1 g E)42.0 g Please give step by step slotion with answaer 14A block of ice at0°C is added to a 150. g aluminum calorimeter cup that holds200. g of water at 10°C. If all but 2. g of ice melt, what wasthe original mass of the block of ice?   

A) 38.8 g B) 62.2 g C) 47.6 g D) 31.1 g E)42.0 g Please give step by step slotion with answaer Please give step by step slotion with answaer

Explanation / Answer

Since all but 2 grams of Ice melted, we can assume that thefinal temperature of the mixture is zero degrees. therefore,calculate the heat lost by the aluminium calorimeter cup and theheat lost by the water and add these together for the total heatlost. This must equal the total heat gained by the ice whileit is melting. Since the ice is already at its melting point,all the heat gained by the ice goes into the latent heat needed forstate change. Step one: Heat lost by calorimeter = mass ofcalorimeter(150g)*temperature change of calorimeter(10 degrees)*specific heat of aluminium. YOu will have to look this up inthe chapter. It is given in a table somewherethere. Step two: Heat lost by the water=mass ofwater(200g)*temperature change of water (10 degrees)* specific heatof water (4186 J/kgC if you are doing it in Joules, otherwise lookup in the same table that you found Al) Step three: Add together your answer in steps one andtwo Step four: Set the number you get in step three equal tothe mass of the ice melted (in this case, let it be X)*the latentheat of fusion for water. I don't remember it, but it shouldbe in the book also. It is about 333 Joules per gram. Solve for X and add 2 to it to get the final answer.

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