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Obviously in a simple, classroom style experiment the energy required to increase water temperature is a constant and thus wouldn't change, but what if we made the experiment just slightly more real world?

Imagine the following set up: a typical metal pot filled with a gallon of water with a heat source below it putting out a constant amount of energy. Now since the output of energy and the rate of absorption by the pot is constant (I assume) I think we can phrase the question as such:

Would the time it takes for the water to go from 50C to 60C be any different from the time it takes the water to go from 80C to 90C?

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  • $\begingroup$ So are you finally asking about the time it takes to heat up water at different temperatures or the energy that it costs? $\endgroup$ – Phonon Aug 3 '14 at 19:07
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The question as posed in the title ("heating a volume of water) has a clear answer: with thermal expansion the mass of the water becomes less and so the specific heat goes down.

But in the body of the question you ask about a constant quantity ("a pot of") of water and several effects start to play:

  1. Water evaporates - this means it loses heat. This heat loss rapidly gets worse as you get closer to the boiling point of the water. Thus unless you have an isolated system, it will definitely take longer to heat from 80 to 90 than from 50 to 60 - because heat input is not the only heat term.
  2. Heat transfer: if you have a constant flame (for example) the heat transfer to the pot will be a function of the temperature of the pot - the hotter the pot, the less efficient the transfer.

When you want to boil water efficiently, you do two things: cover the pot (limit loss due to evaporation) and put the heat inside if you can: for example the submerged heater element in electric kettles. Other forms of boilers also put the heat in the middle of the water (think water heaters for homes) so most of the hot gas gets to give off its energy to the water.

But if you have a flame, the best you can hope to to is transfer all it's internal energy to the water - so when the water is hotter a flame is always less efficient.

Very efficient systems use counter flow - the hot air moves left to right, and the water to be heated right to left: in that way the colder gas meets even colder water so when the gas finally is exhausted it has no heat left. Same principle is used in efficient gas furnace for homes, etc.

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  • $\begingroup$ Would you mind elaborating on why the "hotter the pot, the less efficient the transfer"? Is this simply because the heat difference between the two items is smaller and thus transfers more slowly? $\endgroup$ – user56598 Aug 3 '14 at 20:06
  • $\begingroup$ As for the evaporation and heat loss, is this purely an effect of the heat difference between the water itself and the surrounding air? If we assume that the water had been heated from base temperature in the first place, wouldn't the air surrounding the water have already been heated as part of that process, thus drastically diminishing the heat loss? $\endgroup$ – user56598 Aug 3 '14 at 20:10
  • $\begingroup$ Heat loss of water is mostly due to evaporation - partial vapor pressure increases rapidly with temperature. Heat transfer efficiency is a result of heat flow being proportional to $\Delta T$ - less difference = slower transfer so hot air flows past and escapes without transferring its heat. $\endgroup$ – Floris Aug 3 '14 at 20:13

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