# Calculation of Thermometer Time Lag?

A couple of days ago, contributor 'Rogan Parks' posed the question below the page break. I was about to fully answer it when the OP for no apparent reason deleted his question. I find it an interesting question that appears to have no duplicates on P.Se, so I'm posting it here again, under my name.

I came across an interesting problem, and could not figure out how to solve it:

A thermometer is in thermal equilibrium with its ambient environment. The ambient temperature now increases at a constant rate of $$r (Ks^{−1})$$

The thermometer has a specific heat capacity of $$C (JK^{−1})$$ The rate of transfer equals $$m$$ times the temperature difference.

How do I now model the thermometer reading at any point in time? Any help or pointers in the right direction would be much appreciated.

• Don't you need the thermal diffusivity of the thermometer? – probably_someone Dec 10 '18 at 19:37
• No. We're using lump thermal analysis: consider the temperature of the probe to be homogeneous. That's an approximation, of course. – Gert Dec 10 '18 at 20:17
• That allows the use of the much simpler Newton's law of cooling/heating, instead of Fourier's law of heat conduction. – Gert Dec 10 '18 at 20:35
• Why don't you also present you solution so you can properly get my up vote. – Chet Miller Dec 10 '18 at 20:44
• My point was that the constant in Newton's Law of Cooling will be some function of the thermal diffusivity and the geometry of the thermometer. Or are you assuming that you have that number already? – probably_someone Dec 10 '18 at 20:49

## 1 Answer

Firstly we assume the temperature the thermometer is expected to measure to be: $$T_A=A+rt$$.

Let $$Q$$ be the heat content of the thermometer, then with Newton's Law of heating/cooling:

$$\frac{dQ}{dt}=m(T_A-T_T)=m(A+rt-T_T)$$

An infinitesimal increase in temperature of the thermometer means and increase of $$Q$$:

$$dQ=CdT_T$$

Insert the above in the first equation:

$$C\frac{dT_T}{dt}=m(A+rt-T_T)$$

$$CT_T'=mA+mrt-mT_T$$

But this is not a differential (DE) equation that can be solved by separation of variables.

Rewritten as:

$$CT_T'(t)+mT_T(t)=m(A+rt)$$

it becomes clear it is a linear, 1st order, non-homogeneous DE.

As Chester noted in the comments, it can be solved by use of an integrating factor.

Rearranging slightly:

$$T_T'(t)+\frac{m}{C}T_T(t)=\frac{m}{C}(A+rt)$$

The integrating factor is: $$e^{\int (m/C)dt}=e^{(m/C)t}$$

The solution becomes:

$$T_T(t)=e^{-(m/C)t}\int {\frac{m}{C}(A+rt)e^{(m/C)t}dt}+C_1e^{-(m/C)t}$$

With $$C_1$$ the integration constant. Now develop:

$$\int {\frac{m}{C}(A+rt)e^{(m/C)t}dt}$$ $$=\frac{m}{C}\int {(A+rt)e^{(m/C)t}dt}$$ $$=\frac{m}{C}\Big[A\int {e^{(m/C)t}dt}+r\int t{e^{(m/C)t}dt}\Big]$$ $$=\frac{m}{C}\Big[A\frac{C}{m}e^{(m/C)t}+r\Big(\frac{C}{m}t-\frac{C}{m}^2\big)e^{(m/C)t}\Big)\Big]$$ $$=Ae^{(m/C)t}+r\frac{C}{m}te^{(m/C)t}-r\Big(\frac{C}{m}\Big)^2e^{(m/C)t}$$ Multiplying with $$e^{-(m/C)t}$$, all exponentials drop out and we get: $$A+ r\frac{C}{m}\Big[t-\frac{C}{m}\Big]$$ Adding the integration constant: $$T_T(t)=A+ r\frac{C}{m}\Big[t-\frac{C}{m}\Big]+C_1e^{-(m/C)t}$$ For the determination of $$C_1$$, we'll assume that $$T_T(0)=A$$: $$A=A-r\frac{C}{m}\frac{C}{m}+C_1$$ $$C_1=r\Big(\frac{C}{m}\Big)^2$$ So that: $$T_T(t)=A+ r\frac{C}{m}\Big[t-\frac{C}{m}\Big]+r\Big(\frac{C}{m}\Big)^2e^{-(m/C)t}$$ We could define the lag $$\Delta$$ as: $$\Delta=A+rt-T_T(t)$$ $$\Delta=rt-r\frac{C}{m}\Big[t-\frac{C}{m}\Big]-r\Big(\frac{C}{m}\Big)^2e^{-(m/C)t}$$

$$\Delta=rt\Big(1-\frac{C}{m}\Big)+r\Big(\frac{C}{m}\Big)^2-r\Big(\frac{C}{m}\Big)^2e^{-(m/C)t}$$

$$\Delta=rt\Big(1-\frac{C}{m}\Big)+r\Big(\frac{C}{m}\Big)^2\Big[1-e^{-(m/C)t}\Big]$$

At $$t=0$$ we get $$\Delta=0$$. For $$t \to +\infty \Rightarrow e^{-(m/C)t} \to 0$$, so for $$t\gg 0$$, then:

$$\Delta \approx rt\Big(1-\frac{C}{m}\Big)+r\Big(\frac{C}{m}\Big)^2$$

For large $$t$$ the lag $$\Delta$$ appears to grow linearly.