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If I have a Jacuzzi full of water with $37^{\circ}\text{C}$that is currently not heated within a room full of air with $20^{\circ}\text{C}$, how long will it take for the water (and room) to reach some stable temperature? What temperature will this be for the water (and room)? I assume that there are no other sources of heat/coldness and that the room has no windows or doors (yikes).

If the volume is important, let the Jacuzzi have $500~\text{L}$ of water and the room $50000~\text{L}$ of air

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closed as off-topic by heather, Wolpertinger, ACuriousMind, user259412, sammy gerbil Aug 15 '16 at 19:53

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  • $\begingroup$ @count_to_10 This is a question out of pure personal interest, physics is not my field of study at all but I am willing to learn. If it seems to effortless from my side, I would be grateful just for the answer to the question "which formulas should I look up and understand to solve this question by myself?" $\endgroup$ – IceFire Aug 15 '16 at 12:33
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    $\begingroup$ Fair enough: The water gives heat to the room until a balance is reached. So you need to know how much heat is in the water, then look up specific heat of air, engineeringtoolbox.com/air-specific-heat-capacity-d_705.html which will tell you how much heat it take to raise the temperature of air by 1 degree Celsius. Volumes are vital, to estimate the total heat in the pool and how much air is in the room. Find density of air, to give mass. how long it takes, you will have to Google around for that and look up Newtons Law of Cooling, if you want to be precise. Best of luck with it. $\endgroup$ – user108787 Aug 15 '16 at 12:48
  • $\begingroup$ @count_to_10 Perfect answer to me :) When being new to a field, even using Google seems hard, since it seems difficult to find the correct wording. Most searches resulted in "how long does it take to boil water?" or scientific papers about other particular topics. Your comment gives me plenty of entry point options, though, thank you! $\endgroup$ – IceFire Aug 15 '16 at 12:55
  • $\begingroup$ If the air in the room is maintained at 20 C, I think your intuition would give you a good estimate of the answer to this question. Which do you think it would be closest to: (a) 1 hour, (b) 10 hours, (c) 100 hours? Or choose a range (a) 1 - 10 hours, (b) 10 - 100 hours, or (c) > 100 hours? $\endgroup$ – Chet Miller Aug 15 '16 at 20:27
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In the simplest analysis we start from the principle of energy conservation: heat $Q$ is exchanged between the water and the air in the room, until both are at the same temperature. Heat always flows from hot to cold and stops flowing when the temperature difference between the objects has become zero. We also assume, crucially, that no heat can escape the room.

The thermal energy absorbed (or lost) by an object is given by:

$$Q=mc_p\Delta T$$

Where $m$ is the mass of the object and $c_p$ the specific heat capacity of the object and $\Delta T$ the change in temperature.

I'll use indices $1$ and $2$ to identify the objects and call $T_f$ the final temperature of both objects at thermal equilibrium.

$1$ (the hotter object) loses an amount of thermal energy given by:

$$Q_1=m_1c_{p,1}(T_1-T_f)$$

$2$ (the colder object) absorbs an amount of thermal energy given by:

$$Q_2=m_2c_{p,2}(T_f-T_2)$$

Energy conservation dictates that both quantities must be equal:

$$m_1c_{p,1}(T_1-T_f)=m_2c_{p,2}(T_f-T_2)$$

From the masses, specific heat capacities and initial temperatures the equilibrium temperature $T_f$ can then be calculated.

It has to be noted that this is somewhat idealised because it assumes the masses and specific heat capacities are temperature invariant and that there are strictly no thermal losses. It also assumes there are no other significant objects in the room. Nor does it say anything about how long it will take to reach equilibrium.


As regards the question, 'how long does it take?' I refer to a previous answer, in which I concluded, using Newton's cooling law that the cooling curve, with constant ambient temperature $T_{\infty}$ and $T_0$ the object's initial temperature, is roughly given by:

$$T(t)=T_{\infty}+(T_0-T_{\infty})e^{-\alpha t}$$ Where: $$\alpha = \frac{hA}{mc_p}$$ $h$ is the heat transfer coefficient and $A$ the object's total surface area.

The important lesson to be taken home here is that $T(t)$ follows an exponential decay function. To reach $T_{\infty}$ would in fact take an infinite amount of time!

For some numerical considerations I refer to 'rob's' excellent answer.

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  • $\begingroup$ Yes, jumping the gun like that! ;-) $\endgroup$ – Gert Aug 15 '16 at 13:32
  • $\begingroup$ For a chemist, definitely. :-) $\endgroup$ – Gert Aug 15 '16 at 13:40
  • $\begingroup$ The $\Delta$ notation usually means a difference between two values, e.g. $\Delta T$ means the difference (change) in temperature. It should not be used in combination with Q, in any of your equations. The actual heat exchange is just Q, there is no $\Delta Q$. $\endgroup$ – tvo Aug 15 '16 at 16:53
  • $\begingroup$ Edited the question to briefly answer the 'how long?' question. $\endgroup$ – Gert Aug 15 '16 at 17:31
  • $\begingroup$ There is no such thing as heat content. Closest thing in thermodynamics is 'internal energy' $U$ (or lowercase $u$ per unit mass), which represents energy contained in microscopic kinetic and potential energy modes of e.g. molecules. Heat itself is purely a form of energy EXCHANGE, but can not be contained. Consequently, the $\Delta Q$ notation is meaningless, since $Q$ is not a state variable. Correct is to write $\Delta U = Q = mc_p\Delta T$ in this case. The first '=' is the first law of thermodynamics w/o work, the second '=' uses the definition of heat capacity. $\endgroup$ – tvo Aug 15 '16 at 17:40
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Gert's answer tells you how to compute the equilibrium temperature, but nothing about your title question, which is how fast the cooldown occurs.

That's because computing from first principles how rapidly heat flows from water to air is hard. There are several rules-of-thumb and heuristics that people use, but they're approximations.

When you're insulating a house, you take all the various layers of insulation and add up their thermal resistance to find the rate of heat flow through the walls. The last layer that you include in the calculation is the air film which is found on the outside of the building but is thin enough not to participate in convection. Beyond this distance the assumption is that convection makes the outdoor temperature uniform. This air film thermal resistance is $R=0.17\rm\,h\,ft^2\,^\circ F/Btu$, which seems to correspond to a conductance in SI units of $\rm 35\,W/m^2\,K$.

I don't know whether there's some factor-of-two difference between convection cooling rates from the vertical surface of a wall and convection cooling from the horizontal surface of a pool of water, and I'm ignoring a similar fluid convection layer in the water, so we'll use this as a ballpark thermal conductance.

If your hot tub has a surface area of about 3 m$^2$, your temperature range gives a thermal power of about two kilowatts. Evaporation losses seem to be small. Here's a five kilowatt heating element for a Jacuzzi, so that's the right order of magnitude.

The total heat capacity of your water is $C = mc = \rm 500\,kg \cdot 4.2\,kJ/kg\,K \approx 2100 \,kJ/K$, so drawing power at two kilowatts suggests you should lose about one kelvin every thousand seconds. Note this is an exponential decay, not a constant rate.

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