In the laboratory a \"coffee cup\" calorimeter, or constant pressure calorimeter

Question

In the laboratory a "coffee cup" calorimeter, or constant pressure calorimeter, is frequently used to determine the specific heat of a solid, or to measure the energy of a solution phase reaction. A student heats 69.54 grams of magnesium to 98.74oC and then drops it into a cup containing 83.18 grams of water at 20.64oC. She measures the final temperature to be 33.40oC. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.54 J/oC. Assuming that no heat is lost to the surroundings calculate the specific heat of magnesium.

Explanation / Answer

Heat lost by Mg = mass x specific heat x temperature change

= 69.54 x c x (98.74 - 33.40) = 4543.7436c J


Heat gained by water = mass x specific heat x temperature change

= 83.18 x 4.184 x (33.40 - 20.64) = 4440.8005 J


Heat gained by calorimeter = heat capacity x temperature change

= 1.54 x (33.40 - 20.64) = 19.6504


Total heat lost = total heat gained

4543.7436c = 4440.8005 + 19.6504 = 4460.4509

Specific heat of Mg = c = 0.9817 J/goC = 0.982 J/goC

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