Heating up a water bottle

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  • Thread starter FelixLudi
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Main Question or Discussion Point

Hi,
I didn't know whether or not to put this topic into general physics, classical physics etc. (correct me if I'm wrong on posting here)

Basically I have a normal water bottle, 0.5 L (maybe a little more, up to 0.6 L but definitely not more) and I always heat the water bottle up on a radiator before drinking.

So I have a few questions;
1) Is there a maximum temperature the bottle can reach (not the same as the radiator but lower), before it dissipates more heat rather than it gains?
2) If I fill the water bottle only half way full, will the time required to heat the water up, be exactly half (in regards to heating a full water bottle)?

Considering that the room temperature is average, the bottle is of a cylinder shape (normal shape) with a narrower part at the top, with the water bottle lying only part on top of a radiator (irregularly shaped radiator), and it takes about 1h 30m to heat up the water bottle to "mildly warm" (would call it a solid 30°C (never measured the exact temperature I just drink it when I don't feel it cold in my mouth)).

Disclaimer: This is not a homework question. This is my personal curiosity and interest in opinion and answer from others, as I do not suffice enough knowledge to understand heat dissipation over surface area, heat gain from a source etc.

Note: I will provide any information as it may be required about the question, please do not take this question "jokingly".

Thank you.
 

Answers and Replies

  • #2
jrmichler
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At the beginning, the radiator is putting heat into the bottom of the bottle at a rate proportional to the temperature difference between the radiator and the bottle. As the bottle heats up, the radiator puts less heat into it. If the bottle reached the temperature of the radiator, the radiator would no longer be putting heat into the bottle.

At the beginning, the sides and top of the bottle are either getting heated by the room (if the water is colder than the room) or not losing any heat (if the water is the same temperature as the room). As the water heats up, the sides and top radiate and convect heat into the room. The amount of heat from the bottle into the room increases as the bottle gets warmer.

As the bottle warms, the heat from the radiator into the bottle decreases, while the heat lost from the bottle increases. At some point, the heat added equals the heat lost. At that point, the bottle temperature is stable. It will be less than the radiator temperature (because heat is still going from radiator to bottle) and warmer than the room (because it is losing heat to the room).

Less water in the bottle means it will increase temperature faster because the radiator is putting the same amount of heat into less water.
 
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  • #3
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I agree.
But is the principle of having half a bottle being heated up proportional to having it heat up exactly twice as fast, or will it heat up more than twice as fast because it has less surface area to heat up and it will radiate more but the heater's surface area to the bottle is the same, so it will just add heat more than it loses?
 
  • #4
jrmichler
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Take a look at the equations for free convection heat transfer and radiation heat transfer (Google those terms), along with the assumptions involved. Myself, I'll go along with about twice as fast, but not exactly twice as fast.
 

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