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Is there an equation I can use to calculate the heat required (as a function of time) to sustain water at some specif temperature?

I already know how much heat I need to reach the desired temparature using the formula Q=m*cp *(T2-T1). My question is how much heat do I need to keep 30,000lts of water at 70degress celcius for 30min once i have reached such temperature.

Any help would be greatly appreciated

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    $\begingroup$ You need exactly the same power as is lost from that 30 m³. $\endgroup$
    – Georg
    Commented Dec 1, 2011 at 0:03
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    $\begingroup$ This is highly dependent on what type of Object you are heating your water in. To be more specific, where could the heat escape to? Example, heating in a pot would give you multiple heat-loss opportunities: Heat-Transfer via Air and Radiation (Top of the pot) and via the pot itself. $\endgroup$
    – Michael
    Commented Dec 1, 2011 at 0:15

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If your 30,000 litre tank of water is not losing heat to the surroundings, then the power you need to apply to your tank is zero. While "zero heat loss" might or might not be realistic, it is important to understand this answer for this scenario.

If your 30,000 litre tank of water is losing heat to the surroundings (this could be by conduction through the tank walls, or airflow over an exposed surface (which to richen things up, might include some evaporative effects too)), then to hold the temperature of the tank constant you need to exactly replace the heat that is being lost.

It could be that what you really want to know is "how do I estimate the heat loss from a 30,000 litre tank". This is mostly of an engineering question, your best way to get a good answer might be to consult some engineering handbooks. The answer will depend on the size and shape of the tank, the conductivity of the tank's walls, whether or not the top is open or closed, whether the air surrounding the tank is still (so the heat transfer to the air is via natural convection only) or whether it is in motion (exposed to wind, fans, etc).

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  • $\begingroup$ No need to do such inaccurate calculations. If there is a heater with enough power to raise the temperature to 70 degrees, this heater will be strong enough to maintain that temperature. By watching the on/off ratio of the control circuit, the power can be measured easily and accurate. $\endgroup$
    – Georg
    Commented Dec 1, 2011 at 21:22
  • $\begingroup$ could be accomplished by a feedback control system, which takes the temperature as feedback from system and switch on the heater when temperature reaches a lower water mark. $\endgroup$
    – Vineet Menon
    Commented Jan 31, 2012 at 7:27
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The amount of energy you need to maintain the temperature of water depends on how fast the heat dissipate in your system.

So in your case, if you use a tank to keep 30,000 lts of water, you should take the heat conduction from water to tank into consideration. If you assume the heat conducted from water to tank in a certain amount of time is $H$, then you need to use a heater that has a power of

$P_{heater} = \frac{H}{t}$

to keep the water warm.

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