The Energy Contained in a FunyunI must be mistaken?

  • Thread starter Thread starter Rhine720
  • Start date Start date
  • Tags Tags
    Energy
AI Thread Summary
The discussion revolves around calculating the energy content in a Funyun using a homemade calorimeter, resulting in a measurement of 1018.7 calories for one Funyun. The confusion arises from the distinction between calories and kilocalories, with the calculation yielding 0.848 calories per gram, which likely refers to kilocalories. Participants clarify that food labels use a capital "C" to denote kilocalories, meaning the Funyun's energy content is more plausible at 1.018 kilocalories rather than 1018 calories. The conversation highlights the importance of understanding calorie conversions and the typical daily caloric intake for individuals. Overall, the calculations and terminology used create confusion regarding the actual caloric content of Funyuns.
Rhine720
Messages
88
Reaction score
0

Homework Statement


I'm required to find the amount of energy via calories in a single funyun and then calculate the amount of calories per gram.


Homework Equations


Q=(M)C(deltaT)(Energy gained)
Q=M(deltaT) (Get Q in calories minus C because specific eat for water is same as joules per calory)
Calories per gram = calories /mass of food/1000


The Attempt at a Solution


So I set up my home made Calorymeter(Had to make it) . So I lit it on fire and got managed to get a Delta Temperature. Then did Q=61gx(16.7) And got 1018.7Calories. But my problem is..Can a single Funyun contain 1000+ calories? Am I doing something wrong? I got 0.848 for calories per gram, which sounds reasonable..I may be right but I'm confused about this. Every funyun we eat contains 1k calories?..K i just multiplied 0.848 x 1.2 and got 1.0716 ..Does this have something to do with the bag having a Capital C for 1k calories and what not?
 
Physics news on Phys.org
Food labels have kilocalories on them. 1 Kcal = 1000 calories = 4.2 KJ.
1 kcal for each funyun seems plausible, maybe even too low. Pure fat is about 9 kcal/g. Carbohydrates 4 kcal/g. Whats the weight of a funyun?

You have 1018 Calories for each funyun, but .848 calories per gram. The second number must be kilocalories.
 
I'm did not weigh the funyun. =/ However If those most a healthy person is to eat in a day is 2000-300o calories then how does can a 1018 calory funyun even be realistic?
 
Rhine720 said:
I'm did not weigh the funyun. =/ However If those most a healthy person is to eat in a day is 2000-300o calories then how does can a 1018 calory funyun even be realistic?

Of course those are 2000-3000 kilocalories as well. so if a funyn really has 1018 calories is 1.018 kilocalories, you could eat 2000 a day.
 
Rhine720 said:

The Attempt at a Solution


So I set up my home made Calorymeter(Had to make it) . So I lit it on fire and got managed to get a Delta Temperature. Then did Q=61gx(16.7) And got 1018.7Calories. But my problem is..Can a single Funyun contain 1000+ calories? Am I doing something wrong? I got 0.848 for calories per gram, which sounds reasonable..I may be right but I'm confused about this. Every funyun we eat contains 1k calories?..K i just multiplied 0.848 x 1.2 and got 1.0716 ..Does this have something to do with the bag having a Capital C for 1k calories and what not?
Were you actually instructed to find the calories this way? Could you not just read the Calories and weight info from the package?

Also (as willem2 said) 1 food Calorie = 1000 calories

Note the uppercase "C" is used for food Calories.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top