This article is translated from a Chinese article on my Zhihu account. The original article was posted at 2019-03-23 12:49 +0800.

A field construction requires heating an aluminum alloy component of mass 4.2 kg4.2\,\mathrm{kg} from 10∘C10^\circ\mathrm{C} to 63∘C63^\circ\mathrm{C} or above. Now, we only have 1.2 kg1.2\,\mathrm{kg} of water of 90∘C90^\circ\mathrm{C} in the thermos. Ignoring heat dissipation, design a method to heat the component.

These texts are adapted from some competition problem from Zhejiang. Just throwing the component into the water is not the correct method. The correct answer is to divide the hot water into 33 (or more) parts of equal mass, and heat the components for 33 times.

The more parts we separate the hot water into, the higher the final temperature of the component is. This can trigger our thoughts: what is the final temperature if the number of parts tends to infinity?

Suppose the heat capacity of the component is C0C_0, the initial temperature is T0T_0; the heat capacity of hot water is CC, the temperature is TT. Separate the hot water into nn parts of equal heat capacity (i.e. equal mass). Suppose the temperature of the component is TmT_m after reaching thermal equilibrium with the mmth part of the water. Then, the final temperature is TnT_n.

We now want to find the final temperature at n→∞n\to\infty:

T′≔lim⁡n→∞Tn.T'\coloneqq\lim_{n\to\infty}T_n.

According to the formula of thermal equilibrium,

C0(Tm+1−Tm)+Cn(Tm+1−T)=0.C_0\left(T_{m+1}-T_m\right)+\frac Cn\left(T_{m+1}-T\right)=0.

Solve for Tm+1T_{m+1}:

Tm+1=C0Tm+CnTC0+Cn.T_{m+1}=\frac{C_0T_m+\frac CnT}{C_0+\frac Cn}.

Let k≔C0C0+Cnk\coloneqq\frac{C_0}{C_0+\frac Cn}, and b≔CnTC0+Cnb\coloneqq\frac{\frac CnT}{C_0+\frac Cn}, and we have

Tm+1=kTm+b.T_{m+1}=kT_m+b.

Let m≔0,1,…,n−1m\coloneqq0,1,\ldots,n-1 respectively, and we have

T1=kT0+b,T2=kT1+b,⋮Tn=kTn−1+b.\begin{align*} T_1&=kT_0+b,\\ T_2&=kT_1+b,\\ &\vdots\\ T_n&=kT_{n-1}+b. \end{align*}

Transform the nn equations a little, and we have

kn−1T1=knT0+kn−1b,kn−2T2=kn−1T1+kn−2b,⋮Tn=kTn−1+b.\begin{align*} k^{n-1}T_1&=k^nT_0+k^{n-1}b, k^{n-2}T_2&=k^{n-1}T_1+k^{n-2}b, &\vdots\\ T_n=kT_{n-1}+b. \end{align*}

Sum up all the nn equations, and we have

Tn=knT0+∑j=0n−1kjb.T_n=k^nT_0+\sum_{j=0}^{n-1}k^jb.

According to the formula for summation of geometric progression, we have

Tn=knT0+1−kn1−kbT_n=k^nT_0+\frac{1-k^n}{1-k}b

(this is actually the formula for the nnth recursion of linear function.)

Because

b≔CnTC0+Cn=C0C0+CnCnTC0=C0C0+Cn(C0+CnC0−1)T=(1−C0C0+Cn)T,\begin{align*} b&\coloneqq\frac{\frac CnT}{C_0+\frac Cn}\\ &=\frac{C_0}{C_0+\frac Cn}\frac{\frac CnT}{C_0}\\ &=\frac{C_0}{C_0+\frac Cn}\left(\frac{C_0+\frac Cn}{C_0}-1\right)T\\ &=\left(1-\frac{C_0}{C_0+\frac Cn}\right)T, \end{align*}

then

Tn=knT0+1−kn1−k(1−k)T=T+kn(T0−T).\begin{align*} T_n&=k^nT_0+\frac{1-k^n}{1-k}\left(1-k\right)T\\ &=T+k^n\left(T_0-T\right). \end{align*}

Now calculate the limit of knk^n. We have

lim⁡n→∞kn=lim⁡n→∞(C0C0+Cn)n=lim⁡n→∞1(C0+CnC0)n=lim⁡n→∞1(1+CC0n)n=e−CC0.\begin{align*} \lim_{n\to\infty}k^n&=\lim_{n\to\infty}\left(\frac{C_0}{C_0+\frac Cn}\right)^n\\ &=\lim_{n\to\infty}\frac1{\left(\frac{C_0+\frac Cn}{C_0}\right)^n}\\ &=\lim_{n\to\infty}\frac{1}{\left(1+\frac{\frac C{C_0}}{n}\right)^n}\\ &=\mathrm e^{-\frac C{C_0}}. \end{align*}

Therefore,

T′≔lim⁡n→∞Tn=T+e−CC0(T0−T).\begin{align*} T'&\coloneqq\lim_{n\to\infty}T_n\\ &=T+\mathrm e^{-\frac C{C_0}}\left(T_0-T\right). \end{align*}

This is a intuitive result.

(Gonna take exams for high school entrance…)