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Assuming the following: An isolated space, 1 atm pressure throughout, standard surface tension, vapour pressure and no external force like gravity, Can we provide enough impulse to a drop of water (or a bulk of water) at 300K such that it boils due to the sheer kinetic energy? and if so, how much is required?

Would it be such that we provide kinetic energy equal = latent heat + specific heat * (373-300). I assume that is false but honestly, I have no clue, and not really an expert on the topic of surface tension and vapour pressure.

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    $\begingroup$ What do you think and why? $\endgroup$
    – Bob D
    Commented Mar 27 at 16:43
  • $\begingroup$ If you shoot a bullet into a body of water, the rate of deformation of the water adjacent to the bullet will be sufficient to produce localized boiling, especially if the water is close to 100 C to begin with. $\endgroup$
    – Chet Miller
    Commented Mar 27 at 17:10
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    $\begingroup$ @ChetMiller ooohhh... so can you please help me with some calculations, on how much energy actually will be required. That would be very helpful of you $\endgroup$
    – Le nerd
    Commented Mar 27 at 17:22
  • $\begingroup$ For a bullet, it’s a pretty complicated calculation. $\endgroup$
    – Chet Miller
    Commented Mar 27 at 17:46
  • $\begingroup$ @ChetMiller i mean for my question... where we are just providing an impulse to water $\endgroup$
    – Le nerd
    Commented Mar 27 at 18:10

2 Answers 2

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Boiling Water with Kinetic Energy

Yes, in theory, you could provide enough kinetic energy to a drop of water (or bulk water) at 300K (27°C) to boil it. However, it wouldn't be simply the sum of latent heat and specific heat needed for boiling at standard pressure.

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  • $\begingroup$ "Yes, in theory" well, more than theory. Asteroid impacts are some great examples. $\endgroup$
    – David S
    Commented Mar 27 at 23:25
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Vaporizing water is a two step process:

  1. Heat the water from room temperature to the boiling point: $Q = mc\Delta T$
  2. Vaporize the water at 100°: $Q_v= m H_v$ where $H_v$ is the heat of vaporization for water at 100°

The total heat required to vaporize the water from room temperature is $Q_T = Q + Q_V$

The relationship of heat to kinetic energy is: $$\frac 1 2 mv^2 = Q_T$$ Assuming all the kinetic energy goes to heat.

For one gram of water:

  • $Q = 314.25 J$
  • $Q_V = 2260 J$ Note that the heat required to vaporize 100° water is 7 times larger than the heat required to raise the temperature from 25 to 100°

Vaporizing 1 gram of water is equivalent to one gram traveling at 2,269 m/s. A gram traveling at 793 m/s is equivalent to the energy required to heat the gram of water from 25° to 100°

This is a very simple analysis. The percentage of kinetic energy that goes into thermal energy in a real experiment will be much less than 100%.

If the heat is to be generated due to a collision, then most of kinetic energy goes into conserving momentum.

In a perfectly elastic collision, all the kinetic energy is used to conserve momentum and there is no heat generated.

Inelatic collisions have a portion of kinetic energy that conserves momentum and a portion used to generate heat. The portion used to generated heat must have a kinetic energy of at least $Q_T$ in order to vaporize the water or Q to heat it from 25° to 100°.

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    $\begingroup$ Additional to Stevan V Saban's answer, just having the water traveling at 2269m/s wouldn't vaporize it. You need something, such as a collision perhaps, that would convert that kinetic energy to heat. Even in a collision, some of the water would be vaporized and some would splash away. (btw, Latent heat values have to take into account surface tension etc.) $\endgroup$
    – Rich
    Commented Mar 27 at 20:57
  • $\begingroup$ @Rich Thanks. I've update my answer to include your comments. $\endgroup$
    – Stevan V. Saban
    Commented Mar 27 at 22:49

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