Newton's Law of Cooling
You pull a mug of coffee (90 °C) from the microwave into a 20 °C room. After 5 minutes it reads 70 °C. At what time will it hit a drinkable 60 °C? Newton's Law of Cooling turns this everyday puzzle into a straightforward differential equation — and a formula you can solve in two lines.
A cup of coffee cools from 90 °C to 70 °C in 5 minutes in a 20 °C room. Without any formula — estimate how many more minutes it will take to cool from 70 °C to 50 °C. Write your reasoning below.
Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between the object's temperature and the surrounding (ambient) temperature:
Let $T(t)$ be the temperature of the object and $T_s$ be the constant surrounding temperature. Then:
$$\frac{dT}{dt} = -k(T - T_s), \quad k > 0$$
The negative sign ensures: if $T > T_s$ the object cools ($dT/dt < 0$); if $T < T_s$ it warms ($dT/dt > 0$).
Key facts
- $\dfrac{dT}{dt} = -k(T - T_s)$ with $k > 0$
- Solution: $T(t) = T_s + (T_0 - T_s)\,e^{-kt}$
- $T \to T_s$ as $t \to \infty$ regardless of initial temperature
Concepts
- Why the rate of cooling slows as the object nears ambient temperature
- How $k$ is determined from a second data point using logarithms
- The physical meaning of $T_s$ as the long-run equilibrium
Skills
- Derive the cooling formula from the DE using separation of variables
- Find $k$ given two temperature readings
- Determine temperature at a given time, or time for a given temperature
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.
Quick check: An object cools in a 25 °C room. At $t=0$ it is 85 °C; at $t=10$ it is 65 °C. Which expression correctly gives $k$?
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:
- Identify $T_0$, $T_s$, and any given data point $(t_1, T_1)$.
- Write the solution $T(t) = T_s + (T_0 - T_s)e^{-kt}$.
- Find $k$ by substituting $(t_1, T_1)$ and taking $\ln$.
- 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.
Did you get this? True or false: according to Newton's Law of Cooling, an object will eventually reach exactly $T_s$ in finite time.
Worked examples · 3 in a row, reveal as you go
A metal bar at 200 °C is placed in a 30 °C room. After 10 minutes it cools to 120 °C. Find its temperature after 25 minutes.
Using the same bar from Problem 1 ($T_s = 30$, $T_0 = 200$, $k \approx 0.0637$), find when the bar first reaches 50 °C. Give your answer to the nearest minute.
A cold drink at 4 °C is left in a 28 °C room. After 20 minutes it warms to 14 °C. Find the temperature after 60 minutes and find when it reaches 24 °C.
Fill the gap: An object cools from $T_0$ in a room at $T_s$. The excess temperature $T - T_s$ follows the rule $T - T_s = (T_0 - T_s)e^{-}$$.$
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: Newton's Law of Cooling applies only to objects that are warmer than their surroundings.
Activities · practice with the ideas
Write Newton's Law of Cooling as a differential equation for an object cooling in a 15 °C room, and state the general solution.
A cup of tea at 95 °C cools to 75 °C in 5 minutes in a 25 °C room. Find $k$ exactly (as a logarithm).
Using your answer to Activity 2, find the temperature of the tea after 20 minutes. Give an exact answer then a decimal approximation.
Using Activity 2, find the time (to the nearest minute) at which the tea cools to 40 °C.
Explain in one sentence why the excess temperature $T - T_s$ (not $T$ itself) decays exponentially, with reference to the differential equation.
Odd one out: Three of these statements about Newton's Law of Cooling are correct. Which one is NOT?
Earlier you estimated how long the coffee (90 °C → 70 °C in 5 min, room 20 °C) would take to drop to 50 °C.
From the hook calculation, $k \approx 0.0673$ and $T = 50$ °C is reached at $t \approx 12.6$ min — so about 7.6 more minutes after the 5-min reading. This is longer than the first 5 minutes (90 → 70) because the temperature difference has shrunk from 70 °C to only 50 °C, so cooling slows. This is the heart of Newton's Law: the closer you get to ambient, the slower you cool.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Write Newton's Law of Cooling as a DE and state its solution, defining all symbols. (2 marks)
Q2. A thermometer at 0 °C is placed in a 100 °C room. After 4 minutes it reads 40 °C. Find the temperature after 10 minutes to the nearest degree. (3 marks)
Q3. A pie at 180 °C is removed from an oven into a 22 °C kitchen. After 15 minutes it cools to 120 °C. Find (a) the exact value of $k$, and (b) the time for the pie to reach 60 °C, to the nearest minute. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $\dfrac{dT}{dt} = -k(T-15)$; $T(t) = 15 + (T_0-15)e^{-kt}$ · 2. $e^{-5k} = \dfrac{75-25}{95-25} = \dfrac{50}{70} = \dfrac{5}{7}$; $k = \dfrac{1}{5}\ln\!\left(\dfrac{7}{5}\right)$ · 3. $T(20) = 25 + 70\!\left(\dfrac{5}{7}\right)^4 = 25 + 70 \times \dfrac{625}{2401} \approx 25 + 18.2 \approx 43.2$ °C · 4. $25 + 70e^{-kt} = 40 \implies e^{-kt} = \dfrac{15}{70} = \dfrac{3}{14}$; $t = \dfrac{5\ln(14/3)}{\ln(7/5)} \approx 26$ min · 5. $u = T - T_s \implies \dfrac{du}{dt} = -ku \implies u = Ce^{-kt}$, i.e. the substitution reduces the DE to simple exponential decay.
Q1 (2 marks): $\dfrac{dT}{dt} = -k(T-T_s)$, $k>0$ [1]. $T(t) = T_s + (T_0 - T_s)e^{-kt}$ where $T_0 = T(0)$, $T_s$ = ambient temp, $k$ = positive cooling constant [1].
Q2 (3 marks): $T(t) = 100 - 100e^{-kt}$. $T(4)=40$: $e^{-4k} = 60/100 = 3/5$ [1]. $k = \frac{1}{4}\ln(5/3) \approx 0.1277$. $T(10) = 100-100e^{-10k} = 100 - 100(3/5)^{5/2} = 100 - 100 \times \frac{9\sqrt{15}}{25\sqrt{5}} \approx 100 - 100 \times 0.4665 \approx \mathbf{53}$ °C [2].
Q3 (3 marks): $e^{-15k} = \frac{120-22}{180-22} = \frac{98}{158} = \frac{49}{79}$. (a) $k = \dfrac{1}{15}\ln\!\left(\dfrac{79}{49}\right)$ [1]. (b) $22 + 158e^{-kt} = 60 \implies e^{-kt} = \frac{38}{158} = \frac{19}{79}$; $t = \dfrac{\ln(79/19)}{k} = \dfrac{15\ln(79/19)}{\ln(79/49)} \approx \dfrac{15 \times 1.426}{0.477} \approx \mathbf{45}$ min [2].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering cooling questions. Lighter alternative to the boss.
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