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
