MHT-CET Maths · Differential Equations

Newton's Law of Cooling

A hot body cools at a rate proportional to how much hotter it is than its surroundings. This single named model turns every cooling question into: subtract the surrounding temperature, then track how that difference decays.

All Differential Equations notes MHT-CET Maths strategy

Why this matters

A compact, high-yield model — 5 PYQs sit here (3 HARD, 2 MODERATE) and MHT-CET repeats it almost verbatim year on year (2023, 2024, 2025). Every question is the same shape: cooling data over one interval fixes the rate, and you predict the temperature (or the time) over a second interval. The traps are always the same three: forgetting to subtract the surrounding temperature before taking logs, mishandling the minus sign, and missing the clean (ratio) shortcut when the time-steps are equal.

Concept 1 of 4

The Cooling Model — Rate Proportional to Temperature Excess

Intuition

A cup of coffee cools fast when it is much hotter than the room and slowly once it is nearly room temperature. Newton's law captures exactly that: the cooling rate is proportional to the gap between the body and its surroundings. Once the gap is zero, cooling stops. The key move is to always work with the DIFFERENCE from the surrounding temperature, never the raw temperature.

Definition

Let $\theta$ be the body's temperature and $\theta_s$ the (constant) surrounding temperature. Newton's law of cooling states that the rate of cooling is proportional to the temperature excess $\theta - \theta_s$ :

\dfrac{d\theta}{dt} = -k(\theta - \theta_s), \qquad k > 0.

The minus sign is built in because a body hotter than its surroundings cools DOWN — $\theta$ decreases, so $\tfrac{d\theta}{dt}<0$ .
$k>0$ is a positive constant fixed by the body and medium.
The equation is separable: everything in $\theta$ goes with $d\theta$ , everything in $t$ with $dt$ .

Newton's law of cooling

\dfrac{d\theta}{dt} = -k(\theta - \theta_s), \qquad k > 0

$\theta$ temperature of the body at time t
$\theta_s$ surrounding (ambient) temperature — constant
kpositive cooling constant

Worked example

Water at

90^\circ\text{C}

sits in a room at

20^\circ\text{C}

. Write the differential equation governing its cooling and identify the temperature excess at the start.

Practice this conceptself-check · 4 quick reps

Always work with the excess $\theta - \theta_s$ , not $\theta$

The quantity that obeys clean exponential decay is the temperature EXCESS

\theta - \theta_s

, not the raw temperature

\theta

. A body at

90^\circ

in a

20^\circ

room does not decay toward

0

— it decays toward

20

. Subtract the surrounding temperature before doing anything else.

The minus sign and $k>0$ together mean cooling

Write the model as

\dfrac{d\theta}{dt} = -k(\theta - \theta_s)

with

k>0

. The minus sign is what makes a hot body cool. Absorbing the sign into

k

(letting

k<0

) and then also writing a minus is a common double-negative slip.

Concept 2 of 4

Solving the Cooling Equation — Log Form and Exponential Form

Intuition

The cooling equation is separable, so integrating gives a logarithm of the excess. Exponentiating that gives the working formula: the temperature excess starts at its initial value and multiplies by

e^{-kt}

. Both forms are useful — the log form fits the data, the exponential form predicts the answer.

Definition

Separate and integrate $\dfrac{d\theta}{dt} = -k(\theta - \theta_s)$ :

\int\dfrac{d\theta}{\theta - \theta_s} = -\int k\,dt \;\Rightarrow\; \log(\theta - \theta_s) = -kt + c.

Exponentiating and using the initial excess

\theta_0 - \theta_s

t=0

\theta - \theta_s = (\theta_0 - \theta_s)\,e^{-kt}.

The log form $\log(\theta - \theta_s) = -kt + c$ is what you plug the two data points into.
The exponential form $\theta - \theta_s = (\theta_0 - \theta_s)e^{-kt}$ is what you evaluate for the final answer.

Here $\log$ is the natural logarithm.

Log form and its exponential solution

\log(\theta - \theta_s) = -kt + c \qquad\Longleftrightarrow\qquad \theta - \theta_s = (\theta_0 - \theta_s)\,e^{-kt}

$\theta_0$ initial temperature of the body (at t = 0)
cconstant of integration $= \log(\theta_0 - \theta_s)$

Worked example

A body at

70^\circ\text{C}

is in surroundings at

20^\circ\text{C}

. Given

k = \tfrac{1}{10}\log 2

per minute, find its temperature after

10

minutes.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Differential EquationsMODERATE

The rate at which a substance cools in moving air, is proportional to the difference between the temperature of the substance and that of air. The temperature of air is 290 K and the substance cools from 370 K to 330 K in 10 minutes. Then the time to cool the substance up to 295 K is

[Q114 · 23 April Shift I · 2025]

Take the log of the EXCESS, not the temperature

The integral of

\dfrac{d\theta}{\theta - \theta_s}

\log(\theta - \theta_s)

, never

\log\theta

. Feeding the raw temperature into the log (writing

\log 370

instead of

\log 80

) is the most common wrong start on these questions.

Here $\log$ means natural log

Throughout this model

\log = \log_e

. The base cancels out anyway because you always take a ratio of two logs (or a ratio of excesses), so you never actually need its numerical value — but keep the notation consistent.

Concept 3 of 4

Two-Stage Cooling — Fix the Rate, Then Predict

Intuition

Almost every exam cooling question gives you the temperature after ONE interval and asks for the temperature (or time) after a SECOND interval. The recipe is fixed: use the first interval's data to pin down

e^{-k\Delta t}

(you never even need

k

itself), then substitute into the second interval.

Definition

Divide the two exponential-form equations to eliminate the unknown constants. Writing $E(t) = \theta(t) - \theta_s$ for the excess:

\dfrac{E(t_2)}{E(t_1)} = e^{-k(t_2 - t_1)}.

Procedure:

Stage 1: from the given interval, compute the ratio of excesses to get $e^{-k\Delta t}$ (e.g. $\tfrac{60-20}{80-20} = \tfrac{2}{3}$ ).
Stage 2: raise that ratio to the power (new interval $\div$ first interval) and multiply the current excess by it.

You work entirely with the multiplier $e^{-k\Delta t}$ — the value of $k$ never has to be found.

Ratio of excesses over two intervals

\dfrac{\theta_2 - \theta_s}{\theta_1 - \theta_s} = e^{-k(t_2 - t_1)}

$\theta_1, \theta_2$ temperatures at times t₁, t₂
$\theta_s$ surrounding temperature — subtracted from both

Worked example

A body cools from

90^\circ\text{C}

60^\circ\text{C}

in a room at

30^\circ\text{C}

20

minutes. Find its temperature after

40

minutes.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Differential EquationsHARD

If a body cools from

80°C

50°C

in a room temperature of

25°C

in 30 minutes, then the temperature of the body after 1 hour is

[Q135 · 14th May Shift 2 · 2024]

Subtract the surrounding temperature BEFORE forming the ratio

The ratio that stays constant is of the EXCESSES, not the raw temperatures. For

80\to50

in a

25^\circ

room the multiplier is

\dfrac{50-25}{80-25} = \dfrac{25}{55}

, not

\dfrac{50}{80}

. Using the bare temperatures is the number-one error and gives a wrong answer every time.

Match the exponent to the number of equal intervals

If the first interval is

30

min and the target time is

60

min, that is

60/30 = 2

steps, so the multiplier is SQUARED. For

20

minutes after a

5

-minute interval it is

20/5 = 4

steps — the ratio to the FOURTH power. Miscounting the number of steps changes the exponent.

Drill 1 more on two-stage cooling — fix the rate, then predict

Concept 4 of 4

The (Ratio)ⁿ Shortcut for Equal Time-Steps

Intuition

When the target time is a whole-number multiple of the given interval, you can skip logarithms entirely. Over each EQUAL time-step the temperature excess is multiplied by the SAME fixed ratio — so it forms a geometric progression. Just raise the one-step ratio to the number of steps.

Definition

Over equal time-steps of length $\Delta t$ , the excess $\theta - \theta_s$ is multiplied by the constant factor $r = e^{-k\Delta t}$ each step — a geometric sequence:

\theta_n - \theta_s = (\theta_0 - \theta_s)\,r^{\,n}, \qquad r = e^{-k\Delta t}.

Find $r$ from one interval as a ratio of excesses.
After $n$ equal steps, the excess is $(\theta_0 - \theta_s)\,r^n$ ; add $\theta_s$ back for the temperature.

This is exact (not an approximation) and avoids logs whenever the times are commensurate.

Geometric decay of the excess over n equal steps

\theta_n - \theta_s = (\theta_0 - \theta_s)\,r^{\,n}, \qquad r = e^{-k\Delta t}

rone-step ratio of excesses (constant for equal Δt)
nnumber of equal time-steps = total time ÷ Δt

Worked example

A body cools from

70^\circ\text{C}

50^\circ\text{C}

10

minutes; the surroundings are at

30^\circ\text{C}

. Find the temperature after cooling for

30

minutes.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Differential EquationsHARD

A body cools from

100^\circ\text{C}

60^\circ\text{C}

in 15 minutes. If the surrounding temperature is

20^\circ\text{C}

, the temperature of the body after cooling for one hour is

[Q111 · 2nd May Shift 2 · 2023]

Equal steps ⇒ geometric ratio of the EXCESSES

The excess is multiplied by the same ratio each equal step, so it decays geometrically — NOT linearly. Between

100\to60

the drop was

40^\circ

; the next equal step is not another

40^\circ

but a HALVING of the excess (

80\to40\to20\to10\to5

). Treating cooling as a constant per-step drop overshoots badly.

Count n as total time ÷ interval, then raise the ratio to that power

The exponent

n

is the number of equal intervals, not the number of minutes. For a

15

-min interval and a

60

-min target,

n = 60/15 = 4

, so use

r^4

. Plugging

n = 60

(the minutes) instead of

4

(the steps) is a fatal slip.

Drill 1 more on the (ratio)ⁿ shortcut for equal time-steps

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

The Cooling Model — Rate Proportional to Temperature Excess
Newton's law of cooling
$\dfrac{d\theta}{dt} = -k(\theta - \theta_s), \qquad k > 0$
Solving the Cooling Equation — Log Form and Exponential Form
Log form and its exponential solution
$\log(\theta - \theta_s) = -kt + c \qquad\Longleftrightarrow\qquad \theta - \theta_s = (\theta_0 - \theta_s)\,e^{-kt}$
Two-Stage Cooling — Fix the Rate, Then Predict
Ratio of excesses over two intervals
$\dfrac{\theta_2 - \theta_s}{\theta_1 - \theta_s} = e^{-k(t_2 - t_1)}$
The (Ratio)ⁿ Shortcut for Equal Time-Steps
Geometric decay of the excess over n equal steps
$\theta_n - \theta_s = (\theta_0 - \theta_s)\,r^{\,n}, \qquad r = e^{-k\Delta t}$

Watch out for (8)

Always work with the excess $\theta - \theta_s$ , not $\theta$
→ The Cooling Model — Rate Proportional to Temperature Excess
The minus sign and $k>0$ together mean cooling
→ The Cooling Model — Rate Proportional to Temperature Excess
Take the log of the EXCESS, not the temperature
→ Solving the Cooling Equation — Log Form and Exponential Form
Here $\log$ means natural log
→ Solving the Cooling Equation — Log Form and Exponential Form
Subtract the surrounding temperature BEFORE forming the ratio
→ Two-Stage Cooling — Fix the Rate, Then Predict
Match the exponent to the number of equal intervals
→ Two-Stage Cooling — Fix the Rate, Then Predict
Equal steps ⇒ geometric ratio of the EXCESSES
→ The (Ratio)ⁿ Shortcut for Equal Time-Steps
Count n as total time ÷ interval, then raise the ratio to that power
→ The (Ratio)ⁿ Shortcut for Equal Time-Steps

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