Question #178915

A researcher studying the nutritional value of a new candy places a 6.20 g

6.20 g sample of the candy inside a bomb calorimeter and combusts it in excess oxygen. The observed temperature increase is 2.06 ∘C. 2.06 ∘C. If the heat capacity of the calorimeter is 36.70 kJ⋅K−1, 36.70 kJ⋅K−1, how many nutritional Calories are there per gram of the candy?


Expert's answer

Heat gained by the calorimeter is


Q=Δt∗C=2.06oC∗36.7kJK=75.6kJQ=\varDelta t*C=2.06 ^o C *36.7\dfrac{kJ}{K}=75.6 kJ


1 cal is 4.184 J, so


75.6kJ=18069cal75.6kJ=18069 cal


18069cal6.20g=2914.4calg\dfrac{18069cal}{6.20g}=2914.4\dfrac{cal}{g}


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