Newton's Law of Cooling

Start from the DE $\dfrac{dT}{dt} = -k(T - T_s)$ with $T(0) = T_0$. Let $u = T - T_s$; then $\dfrac{du}{dt} = \dfrac{dT}{dt}$ (since $T_s$ is constant). The DE becomes:

Since $u(0) = T_0 - T_s$, we have $C = T_0 - T_s$. Substituting back $u = T - T_s$:

Finding $k$: If you know $T(t_1) = T_1$ for some $t_1 > 0$, substitute and solve:

Start from the DE $\dfrac{dT}{dt} = -k(T - T_s)$ with $T(0) = T_0$. Let $u = T - T_s$; then $\dfrac{du}{dt} = \dfrac{dT}{dt}$ (since $T_s$ is constant). The DE becomes:

Pause, copy the Newton cooling formula derivation: $\frac{dT}{dt}=-k(T-T_s)\Rightarrow T(t)=T_s+(T_0-T_s)e^{-kt}$ with the substitution $u=T-T_s$ into your book.

We just saw the derivation: $\frac{dT}{dt}=-k(T-T_s)$; let $u=T-T_s$, then $\frac{du}{dt}=-ku\Rightarrow u=u_0 e^{-kt}$, so $T(t)=T_s+(T_0-T_s)e^{-kt}$. That raises a question: given a cooling problem with two temperature readings but an unknown $k$, what is the systematic four-step method to find $k$ and then answer any subsequent sub-part? This card answers it → (1) write the formula; (2) substitute reading 1 for $k$; (3) substitute reading 2 to find a second unknown if needed; (4) answer the question.

Every Newton's Law of Cooling problem follows the same four-step structure:

1. Identify $T_0$, $T_s$, and any given data point $(t_1, T_1)$.
2. Write the solution $T(t) = T_s + (T_0 - T_s)e^{-kt}$.
3. Find $k$ by substituting $(t_1, T_1)$ and taking $\ln$.
4. Answer the question, substitute $t$ to find $T$, or substitute $T$ to find $t$.

When asked to find $T$ at some time: substitute $t$ directly.

When asked to find the time for a given $T$: isolate $e^{-kt}$, take $\ln$ of both sides, divide by $-k$.

Every Newton's Law of Cooling problem follows the same four-step structure:

Pause, copy the four-step cooling problem strategy: write $T(t)=T_s+(T_0-T_s)e^{-kt}$; use one data point to find $k$; use a second if required; answer the question into your book.