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.
 A: 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.
