# Can Newton's Law of Cooling be used to find an initial temperature?

Jul 22, 2015

Yes, if you know the room temperature and the temperature of the cooling object at two known times ${t}_{2} > {t}_{1} > 0$ after the initial time ${t}_{0} = 0$.

#### Explanation:

I will do this in the general case. Somebody else may like to include a specific example (with specific numbers).

BTW, the "infinities" in the subscripts below are supposed to be "double oh's" (they're supposed to be the two middle letters in the word "room").

Suppose the steady room temperature is ${T}_{r \infty m}$, the temperature of the cooling object at time ${t}_{1} > 0$ is ${H}_{1}$, where ${T}_{r \infty m} < {H}_{1}$, and the temperature of the cooling object at time ${t}_{2} > {t}_{1} > 0$ is ${H}_{2}$, where ${T}_{r \infty m} < {H}_{2} < {H}_{1}$.

In this situation, Newton's Law of Cooling can be written as giving the temperature $T$ of the object as a function of time $t$ since the object first started cooling in the following way:

$T = f \left(t\right) = {T}_{r \infty m} + C \cdot {e}^{- k \cdot t}$, where the constants $C > 0$ and $k > 0$ must be found from the given data (that $f \left({t}_{1}\right) = {H}_{1}$ and $f \left({t}_{2}\right) = {H}_{2}$).

This gives two equations in two unknowns ($C$ and $k$ are the unknowns): ${H}_{1} = {T}_{r \infty m} + C \cdot {e}^{- k \cdot {t}_{1}}$ and ${H}_{2} = {T}_{r \infty m} + C \cdot {e}^{- k \cdot {t}_{2}}$.

To solve this system of equations, you could solve the first equation for $C$ in terms of $k$ to get $C = \left({H}_{1} - {T}_{r \infty m}\right) \cdot {e}^{k \cdot {t}_{1}}$ and the substitute this into the second equation to get

${H}_{2} = {T}_{r \infty m} + \left({H}_{1} - {T}_{r \infty m}\right) \cdot {e}^{k \cdot \left({t}_{1} - {t}_{2}\right)}$

This equation can then be solved for $k$ by using logarithms:

${e}^{k \cdot \left({t}_{1} - {t}_{2}\right)} = \frac{{H}_{2} - {T}_{r \infty m}}{{H}_{1} - {T}_{r \infty m}}$

$\setminus R i g h t a r r o w k = \frac{\ln \left(\frac{{H}_{2} - {T}_{r \infty m}}{{H}_{1} - {T}_{r \infty m}}\right)}{{t}_{1} - {t}_{2}} = \frac{\ln \left({H}_{2} - {T}_{r \infty m}\right) - \ln \left({H}_{1} - {T}_{r \infty m}\right)}{{t}_{1} - {t}_{2}}$ (the numerator and denominator here would both be negative, but the fraction would be positive)

We can then ultimately solve for $C$ by substituting into $C = \left({H}_{1} - {T}_{r \infty m}\right) \cdot {e}^{k \cdot {t}_{1}}$ to get:

$C = \left({H}_{1} - {T}_{r \infty m}\right) \cdot {e}^{\left(\frac{\ln \left({H}_{2} - {T}_{r \infty m}\right) - \ln \left({H}_{1} - {T}_{r \infty m}\right)}{{t}_{1} - {t}_{2}}\right) \cdot {t}_{1}}$

Finally, you can solve for the initial temperature ${H}_{0}$ as

${H}_{0} = f \left(0\right) = {T}_{r \infty m} + C {e}^{0} = {T}_{r \infty m} + C$

$= {T}_{r \infty m} + \left({H}_{1} - {T}_{r \infty m}\right) \cdot {e}^{\left(\frac{\ln \left({H}_{2} - {T}_{r \infty m}\right) - \ln \left({H}_{1} - {T}_{r \infty m}\right)}{{t}_{1} - {t}_{2}}\right) \cdot {t}_{1}}$

All this would be easier, of course, if we had specific numbers. But this shows that it is possible to do it in general.