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It is a common suggestion amongst greenhouse enthusiasts that a long-burning tea light inside an upturned terracotta pot can be enough to keep frost away from plants. Is this possible?

Instinctively, it feels wrong: in winter, a greenhouse does not retain heat well so I don't feel a candle can output enough heat to be noticeable, see e.g. this.

From what I could find out, a regular tea light has about $550$kJ of energy. The long-burning ones I would guess, are twice as much so let's say $1100$kJ. Regarding power, they can produce $30$ watts of heat according to this source. Let's say the greenhouse is smallish, about $2.5$m$\times 2$m. I'm not sure where to go from here to calculate the candle's effect.

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Quillo
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Ian Cox
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2 Answers2

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"a greenhouse in the winter does not retain heat well"

This is the crucial point: the rate of heat loss will be roughly proportional to the temperature difference between the inside of the greenhouse and the outside. You typically lose through the glass about 6 joule per second per m$^2$ of glass per °C of temperature difference. We write:

$U_{glass} = 6\ \text {W m}^{-2}\ \text {°C}^{-1}$

If the temperature difference is $\Delta T$ and the glass area is $A$, the rate of heat loss, $Q/t$ will be

$$Q/t = U_{glass}\ A\ \Delta T$$

Using the floor-area you have chosen for your greenhouse and assuming a mean height of 2.5 m, I crudely estimate that $A=30\ \text m^2$

A tea-light, it seems, gives out about 32 W (32 J/s). So by putting $Q/t=32\ \text{J/s}$ we can work out the temperature difference that the tea light could maintain between the inside and outside...

$$\Delta T=\frac {Q/t}{U_{glass}A}=\frac{32\ \text W}{6\ \text {W m}^{-2}\ \text {°C}^{-1}\ 30\ \text m^2}=0.2\ \text{°C}$$

This isn't very impressive is it? And, of course, the tea-light may have to burn for several minutes before this equilibrium temperature difference is established. To be fair , the temperature inside the greenhouse won't be uniform, and it's quite possible that plants near the potted tea-light might be preserved from frost even when the outside temperature is a few degrees below zero.

Philip Wood
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This is a heat transfer rate problem.

The rate of heat generated by the combustion minus the rate of heat transferred through the walls to the outside environment. The bigger this number, the warmer it will be inside.

Here is a good example for you to look at:

https://www.greenhousemag.com/article/basic-greenhouse--engineering-calculations/

Ian
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