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I have a small container (100mm X 80mm X 60mm) filled with water at 10° C. The container is made out of Aluminum and is not insulated, and is resting on a wooden table. Room temperature is about 27° C. How do I calculate the time required for the water in the container to reach room temperature?

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    $\begingroup$ With some difficulty, because it will depend on things like the relative humidity (evaporation and condensation on the outside), the thickness of the aluminum container, as well as things like your altitude. $\endgroup$
    – Jon Custer
    Jun 22 '16 at 19:10
  • $\begingroup$ Evaporation will prevent the water from reaching the temperature of the environment unless the relative humidity of the air is 50%. Otherwise, the evaporation of water from the top of the container (with a rate that is STRONGLY dependent on the presence of air flow across the surface) will continue to remove heat while you are trying to heat the water through the sides of the container. See this earlier answer that estimates a steady state temperature 1C below ambient at 50% relative humidity for a similar scenario. $\endgroup$
    – Floris
    Jun 22 '16 at 23:26
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First get data lined up. It seems the container is not small :)

  1. volume of water: $4.80\times 10^{-4}~m^3$
  2. heat transfer surface area (ignore wood supporting face): $0.038m^2$
  3. mass of water: $0.48~kg$
  4. heat capacity of water is 4.2kJ/kg-K

calculate how much heat required for the water to reach room temperature.

$$\bigtriangleup Q=m c (27-10)= 34.3kJ$$

Assuming water can transfer heat fast, we get the water temperature change rate, $$mc\frac{dT_{water}}{dt}=\dot Q$$

Use $10W/m^2-K$ for air natural convection heat transfer, we get the heat transfer rate, $$\dot Q=hA(T_{room}-T_{water})$$

Thus we end up solving this equation, $$mc\frac{dT_{water}}{dt}=hA(T_{room}-T_{water})$$

Solve this equation, we get $$T_water = 27 - 17 e^{-\frac{hA}{mc}t}=27-17e^{-0.000188t}$$.

Calculate this, you would need about 20 hours. But within 6 hours you can get the water temperature above 90% of 27 degrees.

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  • $\begingroup$ Your final equation seems to tell me that the temperature of the water goes down (without limit) over time. Can this really be what you mean? $\endgroup$
    – WillO
    Jun 22 '16 at 23:09
  • $\begingroup$ Nice calculation! So when I'm at work and I drink more than a Liter a day (9hrs), I can just get a plastic bottle instead of sufficticated thermo aluminum one.. since I drink faster than the speed of temperature raising.. :) think about all the milions you can save if the world knew about it and wouldn't buy those expensive bottles. $\endgroup$ Jun 8 at 13:45
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This kind of question is more reliably answered by doing an experiment, rather than a calculation.

The calculation requires you to decide what are the most important mechanisms for heat loss, what formulas to use, how to take account of the shape of the container, then measure all dimensions, find the appropriate data (thermal conductivities and heat capacities, etc)... A whole lot of work with no guarantee that the answer will be as accurate as you expect.

In the usual mathematical model, the rate of heat loss is proportional to the temperature difference between the container and the room. This results in an exponential rise or fall in temperature towards room temperature. According to this model the container never actually reaches room temperature. You have to decide how close is acceptable. The thermometer gives you a natural criterion (the smallest division). Otherwise it is up to you to decide how close is 'close enough'.

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  • $\begingroup$ But we are Theoretical physicists, Leonard... $\endgroup$ Jun 8 at 13:49

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