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For example, if I'd like to maintain a 1kg sphere of water at 50$\unicode{xb0}$C in a room full of air at 25$\unicode{xb0}$C, how much power must I put into the water as heat? Assume the room is large compared to the water.

I've found explanations and calculators for some specific cases. Most of what I'm finding has to do with the energy transfered as the temperature reaches equilibrium, but I don't know how to transfer that information to this question. I'm not even sure what search terms to use in a case like this where energy is being constantly put into the system.

I'd like to get enough of an understanding to understand the cases in which:

  • The materials are different (e.g. steel ball in water, instead of water in air)
  • There is a third material at the boundary (e.g. water in a rubber membrane in air)
  • The shape is different (e.g. a planar interface, or an arbitrary shaped object)
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  • $\begingroup$ The power should be directly proportional to the rate at which energy is transferred between the object and its environment. Knowing thermodynamic properties isn't enough. We need to know something about the coupling between the object and its environment to answer this question. $\endgroup$ – Ian Aug 29 '16 at 21:12
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Assuming convective losses only (a reasonable assumption at these low temperatures) and assuming uniform water temperature, power loss is given by Newton's cooling law:

$$\frac{dQ}{dt}=hA(T_w-T_{air})$$

Where $A$ is the total surface area of the sphere (easy to calculate for a $1\:\mathrm{kg}$ sphere) and $h$ the convection heat transfer coefficient.

Heat engineering websites put the value of $h$ for solid (the water has to be contained in something) to air convection at about $h\approx 20\:\mathrm{Wm^{-2}K^{-1}}$ at that range of temperatures.

With $A=0.048\:\mathrm{m^2}$ we get:

$$\frac{dQ}{dt}=20 \times 0.048 \times 25=24.2\:\mathrm{W}$$

  1. The materials are different (e.g. steel ball in water, instead of water in air)
  2. There is a third material at the boundary (e.g. water in a rubber membrane in air)
  3. The shape is different (e.g. a planar interface, or an arbitrary shaped object)

Water as the surrounding medium increases the value of $h$ with obvious consequences. Solid to liquid heat transfer is higher than solid to gas heat transfer. $h$ values for free convection solid to water in the range of $50 - 3000\:\mathrm{Wm^{-2}K^{-1}}$ are reported.

The containing material seems to have relatively little influence, as long as it's thin and fairly heat conductive.

Spheres have the highest surface to mass ratio of all regular shaped objects. Switching to a non-spherical shape, all other things being equal, always leads to higher $A$ and thus higher heat loss per unit of time.

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Partial answer (too long for a comment): Water has density about $1 g/cm^3$ so that the volume of 1 kg of water is $1000 \, cm^3$. If the water is in a sphere, the radius of the sphere would be $r \approx 6.2 cm$ . If assume the water is at a uniform temperature $T_w$ then according to the Stefan Boltzmann Law, the surface radiates the amount of energy per second as $\sigma T_w^4 \times 4 \pi r^2$ assuming a black body radiator where $\sigma$ is the Stefan Boltzmann constant, $\sigma \approx 5.7 \times 10^{-12} W/(cm^2 K^4)$, and receives the amount of energy per second from the surrounding atmosphere of $\sigma T_{air}^4 4 \pi r^2$. Here, the temperature of the water sphere and the surrounding air are $T_w = 50^{o} C = 323 K$ and $T_{air} = 25^{o} C = 298 K$ respectively. The sphere has a net loss of energy per second of $\Delta P \approx 8 W$.

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    $\begingroup$ Radiative power losses ten to be small at these temperatures. Don't neglect good old convective losses. $\endgroup$ – Gert Aug 29 '16 at 23:37
  • $\begingroup$ @Gert good point $\endgroup$ – jim Aug 30 '16 at 7:13

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