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Kinetic energy food warmer

Image credit: Aytech solar hand crank charger

Darko Savic
Darko Savic Aug 17, 2021
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The idea is to create a hand-crank food warmer. The heat is created by friction from the rotating crank.

I imagine this could be done in one of 2 ways. It could either be a home appliance that looks something like this:

or it could be a small-form, portable version that is made from 2 components:
  • the hand crank, similar to the one featured in the header image
  • A flexible band filled with oil similar to this:

You wrap the band around the can or whatever you want to warm up. Then start cranking. The heat from friction is transfered to the oil which circulates through the band and back into the heater.

You could use it to heat anything. The band just has to be long enough to wrap around stuff.

This was inspired by Samuel's idea.
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General comments

Samuel Bello
Samuel Bello3 years ago
Such a kinetic heating mechanism will not be practical because the amount of human effort that will be expended to heat up the food is going to be too much, even for an athletic person.

For example, if an 80 pound person climbs a flight of stairs that goes up one floor, this person will exert a force of about 360 newtons through a distance of about 4 meters, hence the person will do about 1440 joules of work. This amount of energy is not nearly enough to raise a kilogram of water by one degree Celcius. The specific heat capacity of water is 4200 joules per kilogram per degree (This means it will take 4200 joules of energy to raise the temperature of one kilogram of water by one degree Celsius). If such a mechanism existed it will be impractically laborious. It is a good idea, but physics will not be on our side.
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Darko Savic
Darko Savic3 years ago
Samuel Bello a typical meal is about 250 to 500 grams or so? It's way less dense than half a liter of water. We just need to warm that up from room temperature to 40-50 degrees celsius. It's some work but shouldn't take more than 5-10 minutes of cranking
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Samuel Bello
Samuel Bello3 years ago
Darko Savic If 250g of water, for example, is to be raised from room temperature(25 degrees Celcius) to 40 degrees, the energy required is

4200*0.25*(40-25)= 15750 joules

This is around the amount of energy needed for an 80 pound person to climb up ten floors using a staircase as in the example I gave earlier. Doing an equivalent amount of work will take more than 5 minutes. The energy required will be more if we account for heat loss. That aside, the work that is required will be much more than the energy the meal will provide to the consumer after it is digested.
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Darko Savic
Darko Savic3 years ago
Yes, it seems you are right. Each ml of water takes 1 calorie to increase by 1 degree celsius. So 250 ml of water would take 250 calories per degree. If we need to increase the food temperature by 15 degrees, so that's 3750 calories. Even if the food is half as dense as water, that's still an extreme amount of calories needed.
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