# Homework Help: A problem about Calorie and Joule transformation

1. Sep 22, 2014

### Quarlep

1. The problem statement, all variables and given/known data
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 ?

2. Relevant equations
W=Fx or W=mgh

3. 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.

2. Sep 22, 2014

### Orodruin

Staff Emeritus
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.

3. Sep 22, 2014

### Quarlep

1 calorie is 4,2 J/gC isnt it so I think it will be 8400000 g m^2/s^2

4. Sep 22, 2014

### Orodruin

Staff Emeritus
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).

5. Sep 22, 2014

### Quarlep

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

6. Sep 22, 2014

### Orodruin

Staff Emeritus
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.

7. Sep 22, 2014

### Quarlep

I understand it thanks

8. Sep 22, 2014

### vela

Staff Emeritus
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.

9. Sep 22, 2014

### Orodruin

Staff Emeritus
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.