Name Date Laboratory lnstructor REPORT SHEET EXPERIMENT Calorimetry: Heat of Fus
ID: 916552 • Letter: N
Question
Name Date Laboratory lnstructor REPORT SHEET EXPERIMENT Calorimetry: Heat of Fusion 9 and Specific Heat Part I: Heat of Fusion Mass of empty calorimeter cup Mass of calorimeter cup+water Mass of water Initial water temperature Final water temperature AT of water Calories of heat lost by liquid water (neg, value) Calories of heat required to melt ice (pos. value) Show calculations Mass of calorimeter cup+ water+melted ice Mass of ice that melted Heat of fusion Show calculations Percent error 10l Copyright C 2015 Pearson Education, Inc.Explanation / Answer
Mass of water present initially in the Calorimeter
= (104.800 - 3.418) g = 101.382 g (Wrong Calculation in the Problem)
Decrease of Temperature is (23.2 - 4.1) = 19.1 oC.
Specific Heat of water is 1 cal / g oC.
Hence, heat released is (101.382 * 19.1 * 1) cal = 1936.3962 cal
Mass of melted ice is (191.021 - 104.800) g = 86.221 g
Heat required to raise the temperature from 0 oC to 4.1 oC is
(86.221 * 4.1 * 1) cal = 353.5061 cal
So, heat absorbed for fusion is (1936.3962 - 353.5061) cal = 1582.8901 cal
Hence, latent heat of fusion is calculated as 1582.8901 cal / 86.221 g = 18.36 cal / g.
Literature value of latent heat of fusion is 79.97 cal / g.
The error % is [(79.97-18.36)/79.97] * 100 = 77.04 % !!!!!!!!!!!!!!!!
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