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A character in my world has a superpower that allows them to convert all (or at least 99.99...%) of the thermal energy contained by the particles within a volume into directed kinetic energy, causing everything within that volume to instantaneously gain a certain velocity while being reduced to a temperature of near absolute zero.

I have been using this thermal energy calculator to determine how much thermal energy a given object would hold, which, using a kinetic energy calculator, then lets me determine the acquired velocity. Assuming a 1kg ball of steel at 20C, I input m=1 (mass of the ball), c=420 (specific heat of steel), and deltaT= 293.15 (temperature difference between 20C and absolute zero). Is that all that needs to be done to calculate the total thermal energy within that steel ball, or is there something I am missing?

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    $\begingroup$ Down votes without comment is not only unhelpful, but detrimental to the platform, and frankly lazy. $\endgroup$
    – Gillgamesh
    Commented Jun 14, 2023 at 12:20
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    $\begingroup$ Probably because the Q is only weakly related to worldbuilding (could easily be phrased for Physics), then from there people hopped on the downvote train. $\endgroup$
    – BMF
    Commented Jun 14, 2023 at 12:35
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    $\begingroup$ I didn't down vote, but this isn't a worldbuilding problem (meaning you're not asking for help creating or consistently using a rule of your world). This is little more than asking for help with physics homework. Granted, it is an appropriate use of the hard-science tag, but I'm having trouble conceiving of an answer that you could apply generally, say... to an apple pie. My opinion about why this isn't a WB question could change if you explain why you need it. (There's a reason SciFi authors don't explain everything to gnat's ear worth of detail.) $\endgroup$
    – JBH
    Commented Jun 14, 2023 at 14:32
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    $\begingroup$ Beginning to hate the definition of the hard-science tag in this stack. In every instance its used 99% of the energy is wasted telling the OP how whatever is in question is impossible. One can get that in the relevant science stack. How about directing some of that intelect to extracting something the OP might need to help him in his creation. Also pointing out the irony that building worlds in itself impossible, yet here we are. $\endgroup$
    – Gillgamesh
    Commented Jun 14, 2023 at 15:05
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    $\begingroup$ @BobaFit I believe you're judging the question based on its backstory. The question itself is only asking about how to calculate a number - and that's a perfectly reasonable hard-science question. $\endgroup$
    – JBH
    Commented Jun 14, 2023 at 21:30

2 Answers 2

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The theorem of equipartition of energy states that molecules in thermal equilibrium have the same average energy. Using the theorem, you can derive a formula for the average kinetic energy of particles in an object:

$$KE=N\frac{3}{2}k_{B}T.$$

On average, each particle has energy $\frac{3}{2}k_{B}T$, where $k_B$ is the Boltzmann constant, $1.38\cdot10^{-23}$ $\text{J⋅K}^{-1}$, and $T$ [K] is the average temperature. $N$ is the average number of particles. I say "average" because objects are made of many different kinds of particles with different masses. However, the theorem states that each one has the same energy on average. So, we can calculate $N$ by:

$$N=\frac{mN_{A}}{M_{bar}}$$

where $m$ [g] is the mass of the object, $N_A$ is Avogadro's number, $6.022\cdot10^{23}$$\text{mol}^{-1}$, and $M_{bar}$ [g/mol] is the average molar mass of particles comprising the object:

$$M_{bar} = M_0(P_0) + M_1(P_1) + ... + M_n(P_n)$$

Where $P_i$ is the mass fraction corresponding to each particular element. If the object is a person made mostly of elemental carbon, oxygen, and hydrogen, at 55%, 45%, and 5% by mass respectively, then:

$$M_{bar} = 12.011(0.55) + 15.999(0.45) + 1.008(0.05)$$

Average body temperature of a human is 310 K, and average mass is 70 kg. Plugging the numbers, I get $KE$ of about about 20 kJ. To get a velocity, simply rearrange the equation for kinetic energy:

$$v = \sqrt{2 \frac{KE}{m}}.$$

I get 23.6 m/s.

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  • $\begingroup$ Cheers, excellent answer. It might even work for an apple pie.... ;-) $\endgroup$
    – JBH
    Commented Jun 14, 2023 at 20:30
  • $\begingroup$ Thank you for the interesting and helpful equations and explanation. Using this formula for a 1kg ball of iron (molar mass of 55.85, temperature of 293.15K) gets me a kinetic energy of 43 kJ, assuming I did the math right (my equation: (6.022x10^23/85)(1.5(1.38x10^-23))293.15). That KE then gets me a velocity of 293 m/s. However, when using the method I described in my question (specific heat of 450, change in temperature of 279.15), I get a thermal energy of 125.6kJ, which translates to a velocity of 500 m/s. Both these numbers work great for story purposes, but I'm still interested.. $\endgroup$
    – M S
    Commented Jun 14, 2023 at 21:26
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    $\begingroup$ @MS I'm not very sure. The Theorem of Equipartition of Energy only deals with particles' translational degrees of freedom. It could be that the extra energy is locked up in rotational and vibrational degrees of freedom, which TEE doesn't address (the formula you found is used to calculate chemical heat energy; Thermal energy is usually associated with kinetic temperature). $\endgroup$
    – BMF
    Commented Jun 14, 2023 at 22:35
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    $\begingroup$ There is also energy associated with intermolecular attractive forces that aren't being accounted for. I don't see anything wrong with either approach (yours or mine), as much energy goes out as what goes in, but the discrepancy is pretty large... $\endgroup$
    – BMF
    Commented Jun 14, 2023 at 22:35
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    $\begingroup$ Yeah, if I had to guess (which I am because this subject isn't my strongest), specific heat capacities are empirical figures found through experiments and automatically take into account the intermolecular bond potentials. Equipartition deals only with the kinetic motion of particles. $\endgroup$
    – BMF
    Commented Jun 14, 2023 at 22:43
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If you want "hard-science" then you need to know that it is physically impossible to do what you suggest. The second law of thermodynamics absolutely does not allow it.

The maximum efficiency that heat can be converted to work depends on having a cold reservoir into which to dump the heat. This efficiency is given by the following formula.

Efficiency = $1 - T_c/T_h$

That is, it is 1 minus the ratio of the cold to the hot temperatures. These are expressed in degrees above absolute zero.

Room temperature is round-about 300 K. (That's a quite warm room at 27C.) If you had some object that was, for example, boiling water, that's 373K. You get an efficiency of 1 - 300/373, or about 19%. That 19% applies only while the hot thing is at boiling temperature. As it cools towards room temperature, the efficiency gets worse and worse. So at 50C it is only 7% or so.

If you had some nearby handy extremely cold reservoir, you could possibly dump heat into that. So, for example, if your superguy had a huge dewar of liquid nitrogen. That's about 77 K. Then at the start you could get up to abut 74% efficiency. As the temperature of the target fell, the efficiency of the process falls. Half way to liquid nitrogen temps it is below 40%.

These calculations assume perfect efficiency of whatever engine you might select. They will, of course, lose some of the theoretical efficiency to friction, etc.

So the result is, even if you were prepared to walk around with a huge cryogenic flask full of liquid nitrogen, you can't get more than a small fraction of the heat energy out as work.

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  • $\begingroup$ Readers please note this answer is a Frame Challenge. $\endgroup$
    – JBH
    Commented Jun 14, 2023 at 14:35
  • $\begingroup$ The question asks about "thermal energy", not heat. True, the question does not specify what is the meaning of the ambiguous phrase "thermal energy", but the answer should begin by explaining that it assumes that by thermal energy the question means heat content or enthalpy or whatever. $\endgroup$
    – AlexP
    Commented Jun 14, 2023 at 14:36
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    $\begingroup$ For a steal ball that stays at 1 atmosphere, the enthalpy, thermal energy, and heat content are indistinguishable. $\endgroup$
    – Boba Fit
    Commented Jun 14, 2023 at 14:42
  • $\begingroup$ Thank you, but please note that this is a super power and as such the laws of physics (and therefore the carnot efficiency) do not necessarily apply. The hard-science tag is only in relation to the stated question, which asks after a method for determining the thermal energy contained within an object. $\endgroup$
    – M S
    Commented Jun 14, 2023 at 20:25

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