Up front: I'm trying to be scientifically accurate about heating up a length of steel chain in a campfire to use as a surprise attack tactic in a D&D campaign, but I'm not sure I'm using the correct formula or searching for the correct terminology. I'm trying to calculate how long the chain has to remain in the fire to cause thermal damage when pulled out and flung at the enemy. It would be 10ft of 0.25 inch thick chain, with about 5ft in the campfire and 5ft out (as a grab handle). The section in the fire would have to reach about 75°C to cause 3rd degree burns.

Here is what I have thus far, which I think is wholly inaccurate.

Rate = Conductivity*A(in m^2)*Temp Diff(in °C)/Thickness (in meters)

Rate of Heat Transfer = kA(T1-T2)/d

T1 = 500°C (Campfire)

T2 = 16°C (Ambient air temp of a cave in a forested area)

d = 0.25 inch thick chain

A = an approximation based on dimension listed in link 1.

Rate = 43A(500-16)/0.00635m

Rate = 43A484/0.00635

Rate = A*20812/0.00635

Rate = A*20812/0.00635

Rate = A*3277480.314960629921259842519685

Rate = 0.0193548*3277480.314960629921259842519685

Rate = 63434.975999999999999999999999999 ~ 63435 Watts

From watts (assuming Rate of Heat transfer was the correct formula to start with), I need a formula to calculate time required to reach a specified temperature.

Links for formulas I used:

Physics Classroom

Chain Dimensions

Engineering Toolbox: Thermal Conductivity of Metals

  • $\begingroup$ What's a "D&D campaign"? $\endgroup$ – Gert Jun 21 at 21:55
  • $\begingroup$ Dungeons and Dragons, specifically the 5th edition rule set, which is commonly referred to as D&D 5E. $\endgroup$ – Steven Miller Jun 21 at 23:02

Using a very simple model, we can use Newton's Law of Cooling and Heating to estimate that time $\Delta t$:

$$\Delta t=\frac{\dot{m}c_p}{h\dot{A}}\ln\Big[\frac{T_f-T_0}{T_f-T(t)}\Big]$$ Full derivation (with slightly different symbols), where:

  • $T(t)$ is the temperature you want the chain to reach (e.g. $75\ \mathrm{C}$) and $T_0$ the chain's starting temperature
  • $T_f$ the fire's temperature
  • $\dot{m}$ the specific mass of the chain in $\mathrm{kg/m}$
  • $c_p$ the specific heat capacity of the chain's material
  • $\dot{A}$ the specific surface area of the chain in $\mathrm{m^2/m}$

Finally, $h$ is the convective heat transfer coefficient. You'll find estimates for your case in engineering handbooks and various dedicated webpages.

However, considering the simplicity of this model (it ignores radiative heating e.g.) and the uncertainty on some of the parameters (like $h$), it might be both simpler and more accurate to determine $\Delta t$ by means of simple experimentation.

[comment] So, if I did the math right and used the correct coefficients: m=0.218, cp=0.2196, h=43, A=0.01935, Tfire=900,T0=16,T(t)=80; then delta time is 0.004319 minutes = 0.259 seconds

Firstly, if calculated in the correct $\text{S.I.}$ units, the result would be in seconds ($\mathrm{s}$), not minutes. This is because of the occurrence of the $\text{Watt}$ i.e. $\mathrm{J/s}$ unit in the formula.

Secondly, the value of $c_p=0.2196$ (unit not specified) is incorrect. For steel we find $c_p\approx 444 \mathrm{J/(kg.K)}$

Thirdly, for $h$ we find a value closer to $h\approx 10\mathrm{W/(m^2.K)}$

I have no choice but to assume the values for $\dot{m}$ and $\dot{A}$ are correct.

So for $\Delta t$ we find:

$$\Delta t=\frac{0.218 \times 444}{10 \times 0.01935}\ln\Big[\frac{900-16}{900-80}\Big]$$

$$\boxed{\Delta t \approx 38\mathrm{s}}$$

This value 'feels right'. I had, just eye-balling the problem, indeed expected a value of seconds rather than minutes or hours.

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  • $\begingroup$ So, if I did the math right and used the correct coefficients: m=0.218, cp=0.2196, h=43, A=0.01935, Tfire=900,T0=16,T(t)=80; then delta time is 0.004319 minutes = 0.259 seconds $\endgroup$ – Steven Miller Jun 21 at 23:36
  • $\begingroup$ I'll look at that tomorrow. Very late here, ta. $\endgroup$ – Gert Jun 21 at 23:58
  • $\begingroup$ I've edited the answer to incorporate your comment. Please upvote or approve, unless you have further questions. $\endgroup$ – Gert Jun 22 at 12:43
  • $\begingroup$ That looks like a more reasonable delta T $\endgroup$ – Steven Miller Jun 23 at 0:15
  • $\begingroup$ Yes. The order of magnitude is definitely correct. 100 %. $\endgroup$ – Gert Jun 23 at 0:48

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