Question

In a typical calorimetry experiment (similar to what you did with the combustion of cheese balls), a small bag of Skittles (10.0 g) was combusted and the temperature of the water (1.00 kg) increased by 40.0 oC. How much energy (in calories and Food calories) was released by the combustion of the Skittles

Answer 1

This question is incomplete, the complete question is;

In a typical calorimetry experiment (similar to what you did with the combustion of cheese balls), a small bag of Skittles (10.0 g) was combusted and the temperature of the water (1.00 kg) increased by 40.0 °C.

a) How much energy (in calories and Food calories) was released by the combustion of the Skittles

b) How many small bags of skittles would a 140 lb person have to eat in order to equal the amount of food calories burned by walking a marathon?

Marathon = 26.2 miles, average person burns 0.453 food calories/mile pound)

Answer:

a)

Number of calories = 40000 Cal

Number of food calorie = 40 KCal

b)

41.54 bags of skittles are required

Explanation:

Given that;

mass of water = 1kg = 1000g

change in Temp ΔT = 40°C

specific heat of H₂O = 4.84 = 1 Cal

Now

heat absorbed by the water = MSΔT

= 1000gm × 1 cal × 40°C = 40000 Cal

so

Number of calories = 40000 Cal

(since food calories = 1000 small calories)

so

Number of food calorie = 40 KCal

Next

number of mile lb , if the man runs in the marathon will be;

26.2 × 140

= 3668 mile lb  

0.453 food calorie is burned for 1 mile pound.

So, amount of food calorie required if the man runs in the marathon will be

0.435 × 3668 mile lb

= 1661.60 food calires.

now

40KCal food calorie is released by 10gm.

So, 1661.60 cal will be released from

⇒ (10gm × 1661.60) / 40 = 415.4 gm

now

1 bag = 10gm,

therefore 415.4 gm = 41.54 bags

∴ 41.54 bags of skittles are required