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General Chemistry fourth edition by McQuarrie, Rock, and Gallogly. University Science Books presented by Macmillan Learning.
A researcher studying the nutritional value of a new candy places a 6.70 g
sample of the candy inside a bomb calorimeter and combusts it in excess oxygen. The observed temperature increase is 2.29 ∘C.
If the heat capacity of the calorimeter is 37.60 kJ⋅K−1,
how many nutritional Calories are there per gram of the candy?

Respuesta :

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Answer:

3.07 Cal/g

Explanation:

Step 1: Calculate the heat absorbed by the calorimeter

We will use the following expression.

Q = C × ΔT

where,

  • Q: heat absorbed
  • C: heat capacity of the calorimeter (37.60 kJ/K = 37.60 kJ/°C)
  • ΔT: temperature change (2.29 °C)

Q = 37.60 kJ/°C × 2.29 °C = 86.1 kJ

According to the law of conservation of energy, the heat released by the candy has the same magnitude as the heat absorbed by the calorimeter.

Step 2: Convert 86.1 kJ to Cal

We will use the conversion factor 1 Cal = 4.186 kJ.

86.1 kJ × 1 Cal/4.186 kJ = 20.6 Cal

Step 3: Calculate the number of Cal per gram of candy

20.6 Cal/6.70 g = 3.07 Cal/g

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