• Pixel-art sticker of Vaporon, personifying Latent Heat of Vaporisation, with the caption “Boiling Costs Energy.”

IndexHeat & EntropySticker

Vaporon

Latent Heat of Vaporisation Physics

\[Q = m L_v\]
Regular price $4.99 USD
Regular price Sale price $4.99 USD
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Surface

Flat $6 shipping worldwideMade to order, ships in 2–5 business days

Details

Size3″ × 3″

3″ × 3″ kiss-cut sticker.

Shipping

Flat $6 shipping on every order, worldwide, whatever's in it.

Delivery times & where we ship

Returns & replacements21 days

Every item is printed to order, so we can’t accept returns or exchanges for change of mind, or if you ordered the wrong size or colour.

Print defect, misprint, damage or the wrong item? We’ll send a free replacement, or a refund if you prefer. Report it within 21 days of delivery with a photo; no need to send it back.

Full refund & replacement policy

The physics

Meet Vaporon, a water droplet in a pot who needs a lot of energy to become steam.

When water boils, the energy you add stops raising its temperature. Instead it pulls molecules apart from their neighbours so they can escape as a gas. The energy needed per kilogram is the latent heat of vaporisation, and for water at 100 °C it is about 2.26 × 10⁶ J/kg.

That is a large number. Warming 1 kg of water from 0 °C to 100 °C takes about 4186 × 100 ≈ 4.19 × 10⁵ J, but turning that same kilogram into steam takes about 2.26 × 10⁶ J, roughly 5.4 times as much. This is why a pot takes so long to boil dry.

It works in reverse too. Steam that condenses on your skin gives back all of that energy, which is why a steam burn can be worse than a burn from water at the same temperature.

Evaporation uses the same idea to cool you down. Sweat takes its latent heat from your skin as it evaporates, carrying energy away from your body on a hot day.

Equations

\[Q = m L_v\]
Symbols
SymbolMeaningUnit
\(Q\) energy needed to vaporise the liquid \(\mathrm{J}\)
\(m\) mass of liquid vaporised \(\mathrm{kg}\)
\(L_v\) latent heat of vaporisation \(\mathrm{J/kg}\)