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.
If you struggled, review the linked prerequisite first.
| Property | Value |
|---|---|
| Concept | Differential Equations |
| Chapter | 6, Section 5 |
| Difficulty | Intermediate |
| Time | ~15 minutes |
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:
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!
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.
| 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 |
| 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 $T_s$.
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.
| 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 |
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 $T = 40$: $$40 = 20 + 70e^{-0.0336t}$$ $$20 = 70e^{-0.0336t}$$ $$e^{-0.0336t} = \frac{2}{7}$$ $$-0.0336t = \ln\left(\frac{2}{7}\right) = -1.253$$ $$t = \frac{1.253}{0.0336} \approx 37.3 \text{ minutes}$$
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 $k_B = -0.2$.
Question 1: A hot object in a cool room has temperature $T(t) = 25 + 75e^{-0.1t}$. What is the room temperature?
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?
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?
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.
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) but asymptotically approaches it, never quite reaching equilibrium.
| 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" |
Looking back:
Looking ahead:
Real-world connections:
| Previous | Up | Next |
|---|---|---|
| Half-Life & Decay | Chapter 6 Skills | §6.6 Skills |
Last updated: 2026-01-22