in lab heat capacity with a calorimeter we measured water to be 105.5 grams then

Question

in lab heat capacity with a calorimeter we measured water to be 105.5 grams then with ice added it became179.9 grams. the initial temperature of this water was 21.5 degrees C and then the final was 9.2 degrees celcius. The shot of metal Pb was measured to be 200.2 grams the intial temperature was 85 degrees celcius and then after heating with the calorimeter the final temperature was found to be 12.6 degrees celcius. We are supposed to find the Cp in cal/gdegreesC. and compare to the accepted value for Cp Pb that is .031 cal/gdegreesC

Explanation / Answer

heat lost by water = 105.5 x 4.184 x (21.5 - 9.2) = 5429.368 J

heat gained by ice = 74.4 x 2.05 x 9.2 = 1403.184 J

heat capacity of calorimeter = (5429.368 - 1403.184)/12.3 = 327.332 J/oC

heat lost by metal = 200.2 x Cp x (85 - 12.6) = 14494.48Cp

so,

105.5 x 4.184 x (21.5 - 12.6) = 14494.48Cp

Cp = 0.27/4.2 = 0.06 cal/g.C

So Cp my this method is calulated to be higher than the actual value

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