A problem about Calorie and Joule transformation

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving energy transformation, specifically the conversion of calories to joules in the context of weightlifting. The original poster presents a scenario where a person consumes 2000 kcal and attempts to calculate how many times they can lift a weight of 45 kg to a height of 2 meters using the work-energy principle.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the conversion of energy units from calories to joules and the implications of using different mass units (grams vs. kilograms) in calculations. There are attempts to clarify the definitions of calories and joules, as well as the relationships between them in the context of work done in lifting weights.

Discussion Status

The discussion is ongoing, with participants providing insights into unit consistency and dimensional analysis. Some participants express confusion over the mixing of units and the implications for the calculations, while others suggest that the original poster's interpretation may not align with standard practices in physics.

Contextual Notes

There is mention of the human body's efficiency in energy expenditure during physical activity, indicating that the calculations may not account for real-world factors affecting energy use.

Quarlep
Messages
257
Reaction score
4

Homework Statement


There's a person and he ate a hamburger which its 2000 kcal.And he wants to spend this energy . he will be weightlifting and he will lift 45 kg to 2 meters.How many times he can lift it ?

Homework Equations


W=Fx or W=mgh

The Attempt at a Solution


2000 kcal=2000000 cal and 1 cal is 4,2 joule so 2000000 cal is 8400000 joule.Its equal mgh so
8400000=42000x10x2xA which a is how many time. so we get 10.But my teacher says the answer will be 10000.The difference is I change kg to grams but she didnt it. [/B]
 
Physics news on Phys.org
The SI unit for mass is kg so that is what should go into your equation as long as you use SI units and expect to get an SI unit (such as Joule) out of the equation. 1 J is equal to 1 kg m^2/s^2 so what you have on the left hand side of your equation is 8400000 kg m^2/s^2 and on your left hand side you have 840000A g m^2/s^2. In order to compare the numbers you need to use the same units on both sides.
 
1 calorie is 4,2 J/gC isn't it so I think it will be 8400000 g m^2/s^2
 
No, calorie is a unit of energy and Joule is also a unit of energy. Thus J/gK would have different units. The calorie is defined as the energy required for heating 1 g of water by 1 K and it is around 4.2 Joule. What is 4.2 J/gK is the specific heat capacity of water - or equivalently 1 cal/gK (which is the definition of 1 cal).
 
Then let me clear Q=mcT and W=mgh so If we takes m we get Q/cT=W/gh If we write the unit form cal cal^-1 g C^-1 C we get g which it grams
and the other side kg m^2 s^-2 s^2 m^-1 m^-1 then we get kg.It does not match.and I think its a problem
 
Now you are just mixing the definition of 1 cal into the problem and this is only confusing. There is nothing being heated in this problem. If you have two sides of an equation in different units you will in general have a conversion factor which in the case one side has a unit of grams and the other side a unit of kg will be 1000. The fact remains that 1 cal is approximately 4.2 J, which in turn is equal to 4.2 kg m^2/s^2. If you put the other side in grams you will have to divide that side by 1000 to put it into kg since 1 g = 0.001 kg.
 
I understand it thanks
 
Quarlep said:
Then let me clear Q=mcT and W=mgh so If we takes m we get Q/cT=W/gh If we write the unit form cal cal^-1 g C^-1 C we get g which it grams, and the other side kg m^2 s^-2 s^2 m^-1 m^-1 then we get kg.It does not match. and I think it's a problem
It's not a problem as both g and kg are units of mass. It would be a problem only if you end up with different types of units for each side. In dimensional analysis, you're not going to get equality unless you stick to one set of units.
 
Let me also add one thing to the original problem: The human body does not have 100% efficiency. Generally you will spend more energy per lift than mgh so in the end you may get away with less lifts to burn that amount of energy.

However, 10 is an unreasonable number. If this was the case you would spend an entire lunch doing 10 repetitions of this exercise. If you go to a gym you will see several people doing significantly more than that without having to eat 100 burgers per day.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K