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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.

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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

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

dθdt=−k(θ−θs),k>0.\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 — θ\theta decreases, so dθdt<0\tfrac{d\theta}{dt}<0.
  • k>0k>0 is a positive constant fixed by the body and medium.
  • The equation is separable: everything in θ\theta goes with dθd\theta, everything in tt with dtdt.

Newton's law of cooling

dθdt=−k(θ−θs),k>0\dfrac{d\theta}{dt} = -k(\theta - \theta_s), \qquad k > 0
  • θ\thetatemperature of the body at time t
  • θs\theta_ssurrounding (ambient) temperature — constant
  • kpositive cooling constant

Worked example

Water at 90∘C90^\circ\text{C} sits in a room at 20∘C20^\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 θ−θs\theta - \theta_s, not θ\theta

The quantity that obeys clean exponential decay is the temperature EXCESS θ−θs\theta - \theta_s, not the raw temperature θ\theta. A body at 90∘90^\circ in a 20∘20^\circ room does not decay toward 00 — it decays toward 2020. Subtract the surrounding temperature before doing anything else.

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

Write the model as dθdt=−k(θ−θs)\dfrac{d\theta}{dt} = -k(\theta - \theta_s) with k>0k>0. The minus sign is what makes a hot body cool. Absorbing the sign into kk (letting k<0k<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

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−kte^{-kt}. Both forms are useful — the log form fits the data, the exponential form predicts the answer.

Definition

Separate and integrate dθdt=−k(θ−θs)\dfrac{d\theta}{dt} = -k(\theta - \theta_s):

∫dθθ−θs=−∫k dt  ⇒  log⁡(θ−θs)=−kt+c.\int\dfrac{d\theta}{\theta - \theta_s} = -\int k\,dt \;\Rightarrow\; \log(\theta - \theta_s) = -kt + c.
Exponentiating and using the initial excess θ0−θs\theta_0 - \theta_s at t=0t=0:
θ−θs=(θ0−θs) e−kt.\theta - \theta_s = (\theta_0 - \theta_s)\,e^{-kt}.

  • The log form log⁡(θ−θs)=−kt+c\log(\theta - \theta_s) = -kt + c is what you plug the two data points into.
  • The exponential form θ−θs=(θ0−θs)e−kt\theta - \theta_s = (\theta_0 - \theta_s)e^{-kt} is what you evaluate for the final answer.

Here log⁡\log is the natural logarithm.

Log form and its exponential solution

log⁡(θ−θs)=−kt+c⟺θ−θs=(θ0−θs) e−kt\log(\theta - \theta_s) = -kt + c \qquad\Longleftrightarrow\qquad \theta - \theta_s = (\theta_0 - \theta_s)\,e^{-kt}
  • θ0\theta_0initial temperature of the body (at t = 0)
  • cconstant of integration =log⁡(θ0−θs)= \log(\theta_0 - \theta_s)

Worked example

A body at 70∘C70^\circ\text{C} is in surroundings at 20∘C20^\circ\text{C}. Given k=110log⁡2k = \tfrac{1}{10}\log 2 per minute, find its temperature after 1010 minutes.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 23 April Shift I · Q114Moderate

Example 2 · Differential Equations · Newton's Law of Cooling

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

Take the log of the EXCESS, not the temperature

The integral of dθθ−θs\dfrac{d\theta}{\theta - \theta_s} is log⁡(θ−θs)\log(\theta - \theta_s), never log⁡θ\log\theta. Feeding the raw temperature into the log (writing log⁡370\log 370 instead of log⁡80\log 80) is the most common wrong start on these questions.

Here log⁡\log means natural log

Throughout this model log⁡=log⁡e\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

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−kΔte^{-k\Delta t} (you never even need kk itself), then substitute into the second interval.

Definition

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

E(t2)E(t1)=e−k(t2−t1).\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−kΔte^{-k\Delta t} (e.g. 60−2080−20=23\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−kΔte^{-k\Delta t} — the value of kk never has to be found.

Ratio of excesses over two intervals

θ2−θsθ1−θs=e−k(t2−t1)\dfrac{\theta_2 - \theta_s}{\theta_1 - \theta_s} = e^{-k(t_2 - t_1)}
  • θ1,θ2\theta_1, \theta_2temperatures at times t₁, t₂
  • θs\theta_ssurrounding temperature — subtracted from both

Worked example

A body cools from 90∘C90^\circ\text{C} to 60∘C60^\circ\text{C} in a room at 30∘C30^\circ\text{C} in 2020 minutes. Find its temperature after 4040 minutes.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 2 · Q135Hard

Example 3 · Differential Equations · Newton's Law of Cooling

If a body cools from 80°C80°C to 50°C50°C in a room temperature of 25°C25°C in 30 minutes, then the temperature of the body after 1 hour is

Subtract the surrounding temperature BEFORE forming the ratio

The ratio that stays constant is of the EXCESSES, not the raw temperatures. For 80→5080\to50 in a 25∘25^\circ room the multiplier is 50−2580−25=2555\dfrac{50-25}{80-25} = \dfrac{25}{55}, not 5080\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 3030 min and the target time is 6060 min, that is 60/30=260/30 = 2 steps, so the multiplier is SQUARED. For 2020 minutes after a 55-minute interval it is 20/5=420/5 = 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

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 Δt\Delta t, the excess θ−θs\theta - \theta_s is multiplied by the constant factor r=e−kΔtr = e^{-k\Delta t} each step — a geometric sequence:

θn−θs=(θ0−θs) r n,r=e−kΔt.\theta_n - \theta_s = (\theta_0 - \theta_s)\,r^{\,n}, \qquad r = e^{-k\Delta t}.

  • Find rr from one interval as a ratio of excesses.
  • After nn equal steps, the excess is (θ0−θs) rn(\theta_0 - \theta_s)\,r^n; add θs\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

θn−θs=(θ0−θs) r n,r=e−kΔt\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∘C70^\circ\text{C} to 50∘C50^\circ\text{C} in 1010 minutes; the surroundings are at 30∘C30^\circ\text{C}. Find the temperature after cooling for 3030 minutes.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 2 · Q111Hard

Example 4 · Differential Equations · Newton's Law of Cooling

A body cools from 100∘C100^\circ\text{C} to 60∘C60^\circ\text{C} in 15 minutes. If the surrounding temperature is 20∘C20^\circ\text{C}, the temperature of the body after cooling for one hour is

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→60100\to60 the drop was 40∘40^\circ; the next equal step is not another 40∘40^\circ but a HALVING of the excess (80→40→20→10→580\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 nn is the number of equal intervals, not the number of minutes. For a 1515-min interval and a 6060-min target, n=60/15=4n = 60/15 = 4, so use r4r^4. Plugging n=60n = 60 (the minutes) instead of 44 (the steps) is a fatal slip.
Drill 1 more on the (ratio)ⁿ shortcut for equal time-steps

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Test yourself on Differential Equations

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