# How do we calculate energy of food?

Here is the problem, according to me:

In a classic hamburger, according to my internet research, approximately 906 kJ are contained.

Now take a car of 1 000 kilograms, and push it to 30 m/s. It has a certain kinetic energy, which is: 1 000 * 30² = 900 kJ

So does it mean that the work (in a physical sense) needed to stop the car is equivalent to the energy brought by the hamburger, so a human just have to eat an hamburger to be able to stop the car?

It is very hard to imagine, as even with the strength of an human and the possibility to brake the car very slowly, I think a human will need a lot of energy.

So how it is possible to compare the energy brought by food and the mechanical/kinetic energy?

Because on the other hand, it could also mean that with the energy brought by one hamburger (approximately 15% of the food energy is used by the human body, and the value I gave take into account this 85% factor), I am able to put push a car to 30 m/s!

• youtu.be/49EaMd1fu-E – Gert Aug 23 at 19:06
• " It has a certain kinetic energy, which is: 1 000 * 30² = 900 kJ" The kinetic energy is actually given by $K=\frac12 mv^2$. So it's 450 kJ, not 900. – Gert Aug 23 at 19:11
• Note that strength isn't the same as work. Also, work doesn't specify a time period to perform that work. – Aaron Stevens Aug 23 at 20:04
• "the work (in a physical sense) needed to stop the car" This part might hide a deep misunderstanding of how energy and work work: Generally, it does not require work to stop a car: just crash the car into something, and the something actually gains energy from the car, often in the form of breaking or starting to move. Humans need energy to push a car to a stop because human muscles are not stiff when excerting a force: there's small movement all the time which heats the muscles. But that energy doesn't need to have anything to do with how much energy the car loses by – JiK Aug 24 at 12:51
• @totalMongot Work has a sign. A force on an object in the direction opposite to the movement of the object does negative work on the object. – JiK Aug 24 at 13:19

A calorimeter is a device that you can use to measure the energy of food. You put a small amount of the food in a calorimeter. Then you submerge the calorimeter in a known quantity of water. Now you burn the food and measure the increase in temperature of the water. By careful measurements, one can determine the energy required to heat the water. That amount of heat is the caloric content of that food.

Once you know the energy of the food you can compare it to other material objects just by calculating the energy of the objects. Kinetic energy is just $$(1/2) mv^2$$

• Is it a fact? Humans cannot oxidize substantial parts of food, cows some other parts, whereas this method oxidizes everything. The result won't be very relevant for us. – kubanczyk Aug 24 at 13:18
• @kubanczyk I may be wrong on this, but I guess the process described in this answer is used to determine the amount of energy provided by the digestible constituents of food. I.e. run one test for fat, another for the protein mix in meat, another for the protein mix in wheat, another for the starch in potatoes. Then analyse the nutrient mixture of the food in question and add up the respective energy contents. Thus, indigestible parts like fibers are naturally excluded from the calculation. – cmaster Aug 24 at 14:13
• This; you burn it. Which begs the question that's already answered in it: "approximately 15% of the food energy is used by the human body" – Mazura Aug 24 at 18:43

It isn't so far fetched as it might appear. A gallon of gas will accelerate the car to speed and keep it there for perhaps 30 miles. The energy in gas comes from chemical bonds. The bonds in gas are not totally different from the bonds in a hamburger.

Burning a gallon of gas produces 132,000 BTU, or $$1.39 \times 10^8$$ J. That is ~150 times more than the 900,000 J from burning a hamburger. Given the relative sizes, it sounds like gas has more energy per pound. So a hamburger has enough energy to run a car for about 0.2 miles.

Eating a hamburger for lunch produces enough energy to run a human until dinner. Over the course of an afternoon, a human could do enough work to push the car up to speed. Perhaps he could pedal a bicycle connected to a generator, and store the energy in batteries. Then the batteries might well have enough charge to get an electric car up to speed.

• Interesting. But the way the hamburger is "burnt" in the human body is not the same as a gallon of gas is literrally "burnt" in a motor But my understanding is that there are some bounds more or less stable: gas bounds are less stable than hamburger bounds so more energy could be extracted from. So is there any reason that could be added to say that the energy at the end would not be the same? – totalMongot Aug 23 at 19:47
• It actually isn't that different. In both cases, fuel is combined with oxygen to produce CO2 and H2O. It just happens slower in a human. It produces heat, which keeps our body temperature at 98.6. But not the flame you get from running the reactions faster, or the explosion from faster still. – mmesser314 Aug 23 at 19:52
• But is a human able to concentrate all the heat/energy Not strictly all of it because some is needed for vital functions. But energy is energy is energy.... – Gert Aug 23 at 20:11
• But is a human able to concentrate all the heat/energy Well, is any engine able to convert all the energy into a useful work? – kubanczyk Aug 24 at 13:22
• "Perhaps he could pedal a bicycle connected to a generator [...]" Or he could just forget about the car and use the 900kJ to ride about 7km on his bike... Nothing beats the fuel efficiency of a cyclist. – cmaster Aug 24 at 14:05

There's something you're overlooking that makes the whole thing somewhat counter-intuitive: the maximum force the human can exert on the car.

In order to reach the kinetic energy of $$450\mathrm{kJ}$$ that same amount of work $$W$$ has to be done on the car by the human, according to:

$$W=K$$

where:

$$\mathbf{d}W=F\mathbf{d}x$$

Here $$F$$ is the force exerted by the human (we'll assume it constant, for simplicity's sake) on the car and $$\mathbf{d}x$$ the displacement (linear, for simplicity's sake). Integrating we get:

$$W=F\Delta x$$

and:

$$F\Delta x=\frac12 mv^2$$

Or:

$$\Delta x=\frac{m v^2}{2F}$$

Obviously if $$F$$ is relatively small compared to $$mv^2$$, $$\Delta x$$ will be large. However that doesn't mean the human won't get the car up to speed: it just takes quite a long time.

That time can be calculated from Newton's second Law:

$$F=ma \to a=\frac{F}{m}$$

and:

$$v=a\Delta t \to \Delta t=\frac{v}{a}$$

So:

$$\Delta t=\frac{mv}{F}$$

It's similar if we made the car tow a fully loaded $$20$$ tonnes HGV: the car can do it but it would take an impractical amount of time because the maximum amount of force it can deliver is small compared to that of an HGV tractor.

• I think you should mention friction somewhere. – infinitezero Aug 24 at 8:47
• +1 but it's not only about maximum force, it's also about maximum power. With rope and pulley systems, the problem of maximum force can be avoided, but still we'd have the problem of maximum power. With a good quality and ridiculously big system, a human could actually pull a car to fast speeds, but it still requires time because there's maximum power. But as $P=Fv$, the initial force could be quite big with an ideal rope and pulley system. – JiK Aug 24 at 12:55
• ...and obviously, the inefficiency of human muscles needs mentioning. – JiK Aug 24 at 13:25

The whole scenario seems to be implausible when you consider that in order for the person to accelerate the vehicle to a speed of 30 m/s at the end the person would have to run at a pace of 70 mph!

A more reasonable speed to push the car is about 10 mph or 4.5 m/s which is a kinetic energy of about 10 kJ. That’s only 10% of your hamburger’s calories.

By the way, according to one site an average person's straight pushing force is 60-80 N. At 80 N assuming negligible friction the car's acceleration would be 0.08$$\frac{m}{s^2}$$. So If would take about a minute to reach 4.5 m/s.

Hope this helps.

• You can easily overcome the problem that a human can't run at 30 m/s by changing the problem to a human-powered vehicle. And the "straight pushing force" you quote ignores the fact that the human can use their own weight to apply force to the car. A normally fit person can accelerate a car up to walking pace (say 2 m/s) in a lot less time than 30 seconds. – alephzero Aug 24 at 23:52
• @alephzero Couple of issues here. First, the OP was clearly talking about a person pushing the vehicle, not a human powered vehicle. Second, the force quoted on the site didn't discuss how the person applies it, just that it is horizontal. So it probably includes both the force of leaning into the vehicle, and the force applied due to the friction between the person's feet and the ground. In any case, the amount of force is not the main issue. It's the feasibility of burning the hamburger calories by pushing the car. – Bob D Aug 25 at 16:55

I am able to put push a car to 30 m/s!

You are ignoring the inefficiency of your body burning those 900 kJ here, your actual mechanical output will be much lower.

Ok, so let's assume that you can actually produce 25 % of the burned energy as mechanical work, then, yes, you can accelerate the car to 15 m/s (half the speed, quarter the energy). If the car is suspended frictionlessly in vacuum, that is. It'll take you a while, because you will be feeding this energy into the car at about 200 to 250 watts, so it'll take you about $$225\ \mathrm{kWs}/225\ \mathrm W = 1000\ \mathrm s = 16\ \mathrm{min} + 20\ \mathrm s$$ to get the car up to speed.

Unfortunately, car's loose much more than the 250 watts that you can output to wind resistance when going 54 km/h, so you won't reach that speed without the vacuum. You'll only reach the speed at which the losses to friction and wind resistance match your energy input, which is a rather meager figure. Better get on your bike and ride about 7 km at 25 km/h with the energy your hamburger provided.