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A researcher studying the nutritional value of a new candy places a 6.80 g 6.80 g sample of the candy inside a bomb calorimeter and combusts it in excess oxygen. The observed temperature increase is 2.25 ∘ C. 2.25 ∘C. If the heat capacity of the calorimeter is 32.00 kJ ⋅ K − 1 , 32.00 kJ⋅K−1, how many nutritional Calories are there per gram of the candy?

Respuesta :

Answer:

2523.23 cal / g

Explanation:

Mass = 6.80g

Temperature increase (T2 - T1) = 2.25

Heat capaity of calorimeter = 32.00kJK-1 = 7.64818 KcalK-1 ( Converting to calories by multiplying by 4.18)

Heat capacity = 7648.18 cal (converting to cal by mutltiplying by 1000)

Heat capacity = Heat Supplied / Temperature rise

Heat Supplied = Heat capacity * Temperature rise

Heat supplied = 7648.18 * 2.25

Heat Supplied = 17208.405 cal

To obtain the per gram value, we divide by the mass.

This gives us; 17208.405 / 6.80 = 2523.23 cal / g

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