Newton's Law of Cooling
Why Coffee Cools Faster When It’s Hotter
You’ve noticed it: hot coffee cools quickly at first, then more slowly as it approaches room temperature. A physicist in the 1700s quantified this observation. Newton discovered that the rate of cooling is proportional to the temperature difference between an object and its surroundings.
This simple principle lets us predict when your coffee will be drinkable, determine time of death in forensics, and optimize heating/cooling systems in engineering.
Before You Start
Quick Self-Check
Solve $\frac{dy}{dt} = -0.2y$ with $y(0) = 50$. → Exponential Growth ODE
Solve $e^{-0.3t} = 0.4$ for $t$. → Natural Logarithm
If you struggled, review the linked prerequisite first.
Prerequisite Map
Quick Reference
| Property | Value |
|---|---|
| Concept | Differential Equations |
| Chapter | 6, Section 5 |
| Difficulty | Intermediate |
| Time | ~15 minutes |
Key Concepts
Newton’s Law of Cooling
The rate of temperature change is proportional to the difference between the object’s temperature and the surrounding temperature:
$$\boxed{\frac{dT}{dt} = k(T - T_s)}$$
where:
- $T(t)$ = temperature of object at time $t$
- $T_s$ = surrounding (ambient) temperature (constant)
- $k$ = cooling constant ($k < 0$ for cooling, $k > 0$ for warming)
The Substitution Trick
This doesn’t look like $\frac{dy}{dt} = ky$, but we can transform it.
Let $y(t) = T(t) - T_s$ (the temperature difference).
Then $\frac{dy}{dt} = \frac{dT}{dt}$ (since $T_s$ is constant), and the equation becomes:
$$\frac{dy}{dt} = ky$$
This is our standard exponential model!
The Solution
Since $y(t) = y_0 e^{kt}$ and $y = T - T_s$:
$$T(t) - T_s = (T_0 - T_s)e^{kt}$$
$$\boxed{T(t) = T_s + (T_0 - T_s)e^{kt}}$$
where $T_0 = T(0)$ is the initial temperature.
Understanding the Formula
| Component | Physical Meaning |
|---|---|
| $T_s$ | Temperature the object approaches as $t \to \infty$ |
| $T_0 - T_s$ | Initial temperature difference |
| $(T_0 - T_s)e^{kt}$ | Decaying temperature difference |
| $k$ | How fast the object equilibrates |
Cooling vs Warming
| Scenario | Condition | Sign of $k$ |
|---|---|---|
| Cooling | $T_0 > T_s$ | $k < 0$ |
| Warming | $T_0 < T_s$ | $k < 0$ (same equation!) |
The formula works for both! If $T_0 < T_s$, then $T_0 - T_s < 0$, and the temperature increases toward $T_s$.
Long-term Behavior
As $t \to \infty$: $$\lim_{t \to \infty} T(t) = T_s + (T_0 - T_s) \cdot 0 = T_s$$
The object’s temperature approaches the ambient temperature.
Common Pitfalls
| Mistake | Why It’s Wrong | Correct Approach |
|---|---|---|
| Using $T(t) = T_0 e^{kt}$ | This ignores the ambient temperature shift | Use $T(t) = T_s + (T_0 - T_s)e^{kt}$ |
| Forgetting to subtract $T_s$ when finding $k$ | The data point $T(t_1) = T_1$ must be processed as $(T_1 - T_s) = (T_0 - T_s)e^{kt_1}$ | Always work with temperature differences |
| Thinking temperature can drop below $T_s$ | The exponential approaches 0, so $T \to T_s$ | Temperature asymptotically approaches ambient |
| Confusing cooling ($k < 0$) with the sign of $T_0 - T_s$ | $k < 0$ always for this model; the initial difference determines direction | Keep $k < 0$; sign of $T_0 - T_s$ handles cooling vs warming |
| Wrong ambient temperature | Using room temp when object is in a fridge, for example | Identify $T_s$ carefully from the problem |
Worked Example
Problem: A cup of coffee at 90°C is placed in a 20°C room. After 10 minutes, the coffee has cooled to 70°C.
(a) Find the cooling constant $k$. (b) Find the temperature after 20 minutes. (c) When will the coffee reach 40°C?
Solution:
(a) We have $T_0 = 90$, $T_s = 20$, and $T(10) = 70$.
Using $T(t) = T_s + (T_0 - T_s)e^{kt}$: $$70 = 20 + (90 - 20)e^{10k}$$ $$50 = 70e^{10k}$$ $$e^{10k} = \frac{5}{7}$$ $$k = \frac{\ln(5/7)}{10} = \frac{-0.336}{10} = -0.0336$$
(b) After 20 minutes: $$T(20) = 20 + 70e^{-0.0336 \times 20} = 20 + 70e^{-0.672}$$ $$= 20 + 70 \times 0.511 = 20 + 35.8 = 55.8°\text{C}$$
(c) When $T = 40$: $$40 = 20 + 70e^{-0.0336t}$$ $$20 = 70e^{-0.0336t}$$ $$e^{-0.0336t} = \frac{2}{7}$$ $$-0.0336t = \ln\left(\frac{2}{7}\right) = -1.253$$ $$t = \frac{1.253}{0.0336} \approx 37.3 \text{ minutes}$$
Practice Problems
A thermometer reading 25°C is placed in a refrigerator at 4°C. The cooling constant is $k = -0.15$ per minute. Write the temperature function $T(t)$.
A pie at 180°C is taken from the oven and placed in a 22°C kitchen. The cooling constant is $k = -0.05$ per minute. What is the pie’s temperature after 30 minutes?
A frozen turkey at $-18°$C is placed in a 25°C room. After 2 hours, the turkey’s temperature is 5°C. Find the cooling constant and predict the temperature after 4 hours.
A body is discovered at 10:00 PM with temperature 32°C. At 11:00 PM, the body temperature is 30°C. The room temperature is 20°C, and normal body temperature is 37°C. Assuming Newton’s Law of Cooling, estimate the time of death.
Two objects A and B are placed in the same room at temperature $T_s$. Object A starts at $T_A = 100°$ and B starts at $T_B = 60°$, where $T_s = 20°$. Object A has cooling constant $k_A = -0.1$ and B has $k_B = -0.2$.
- Write temperature functions for both objects.
- Will the objects ever have the same temperature? If so, when?
- Prove that if two objects have the same ambient temperature and the hotter object has a smaller magnitude cooling constant, their temperatures will eventually be equal.
CCI-Style Conceptual Questions
Question 1: A hot object in a cool room has temperature $T(t) = 25 + 75e^{-0.1t}$. What is the room temperature?
Answer
25°. The room temperature is $T_s$, the constant term in the formula. As $t \to \infty$, $T(t) \to 25$.
Question 2: Object A cools from 80°C to 60°C in 10 minutes. Object B cools from 80°C to 60°C in 5 minutes. Both are in the same 20°C room. Which has the larger magnitude cooling constant?
Answer
Object B. It cools faster (reaches 60°C sooner), so $\vert k_B\vert > \vert k_A\vert $.
Question 3: According to Newton’s Law of Cooling, can an object ever cool below the ambient temperature?
Answer
No. The formula $T(t) = T_s + (T_0 - T_s)e^{kt}$ shows that as $t \to \infty$, $T(t) \to T_s$. The object asymptotically approaches room temperature but never goes below it.
Common Misconceptions
in Newton’s law of cooling, the object temperature $T(t)$ eventually reaches zero rather than the ambient temperature $T_s$.
This is the asymptote-as-wall error. The solution $T(t) = T_s + (T_0 - T_s)e^{kt}$ (with $k < 0$) approaches $T_s$ as $t \to \infty$ because $e^{kt} \to 0$, driving the exponential term to zero and leaving only $T_s$. The object cools toward the surrounding temperature, not toward absolute zero. This is also why the model uses $T - T_s$ as the relevant quantity: it is that difference, not the absolute temperature, that decays exponentially.
Mastery Checklist
Mental Model
The “Leaky Bucket” Analogy:
Imagine heat as water in a bucket with a hole. The more water (temperature difference), the faster it leaks out (cools). As the water level drops, the leak slows down. The bucket never quite empties; it approaches the “ground level” (ambient temperature) asymptotically, never quite reaching equilibrium.
Still Confused?
| If you’re struggling with... | Review this |
|---|---|
| The basic exponential model $y' = ky$ | Exponential Growth ODE |
| Solving equations with $\ln$ | Natural Logarithm |
| The substitution $y = T - T_s$ | Think of it as “temperature difference from ambient” |
Connections
Looking back:
- Exponential Growth ODE provides the base model
- The substitution $y = T - T_s$ is a key technique for shifted exponentials
Looking ahead:
- Separable differential equations generalize this technique
- Heat transfer in engineering uses similar principles
- Thermal equilibrium in thermodynamics
Real-world connections:
- Forensic science (time of death estimation)
- HVAC system design
- Cooking and food safety
- Climate modeling
| Previous | Up | Next |
|---|---|---|
| Half-Life & Decay | Chapter 6 Skills | §6.6 Skills |
Last updated: 2026-01-22