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A cup of tea on your desk cools quickly at first, then more and more slowly. Newton's law of cooling turns that observation into one equation. Exam questions then test it in two ways: with the exact exponential law, and with a quicker "average temperature" method that many notes teach. The two methods can give different answers to the same question. This post works one NCERT problem both ways, shows when the shortcut is safe, lists the usual mistakes and ends with a practice routine.
Which exams ask about Newton's law of cooling?
The JEE (Advanced) 2026 brochure names "Newton's law of cooling" in its thermal physics syllabus. The CBSE Physics curriculum for 2026-27 includes a Class 11 experiment "to study the relationship between the temperature of a hot body and time by plotting a cooling curve". The JEE (Main) 2026 syllabus lists heat transfer by conduction, convection and radiation without naming the law. Always check the current year's documents for your exam.
What does the law actually say?
The NCERT chapter Thermal Properties of Matter states that the rate of loss of heat is directly proportional to the temperature difference between the body and its surroundings, and adds that "the law holds good only for small difference of temperature". In symbols, with T the body's temperature and Ts the surroundings:
dT/dt = −k(T − Ts)
Integrating gives T − Ts = (T₀ − Ts)e^(−kt). The quantity that decays is the excess temperature, T − Ts, not the temperature itself. Like radioactive decay, the excess halves in equal time intervals. The NCERT Class 11 physics laboratory manual uses this in Experiment 14: a graph of log₁₀(θ − θ₀) against time is a straight line, with slope equal to −k/2.303.
Why is it only an approximation?
Heat leaves a hot object by convection and by radiation, and radiation does not grow in proportion to the temperature difference. Michael Vollmer showed in Newton's law of cooling revisited (2009) that simple exponential cooling is mostly valid below a threshold temperature difference that depends on the set-up. For larger differences, his infrared measurements on objects up to 300 K hotter than their surroundings showed clear deviations.
You can see the link to Stefan's law yourself. For a small excess ΔT, T⁴ − Ts⁴ ≈ 4Ts³ΔT, so the net radiated power becomes proportional to ΔT, which is Newton's law. Our post on black-body radiation and Stefan's law covers the full version.
Worked example 1: the exact method
NCERT exercise 10.20 asks: a body cools from 80 °C to 50 °C in 5 minutes. How long does it take to cool from 60 °C to 30 °C? The surroundings are at 20 °C.
Convert to excess temperatures. The first cooling takes the excess from 60 to 30. The second takes it from 40 to 10.
The first interval halves the excess in 5 minutes, so the "half-life" is 5 minutes and k = ln 2 / 5 ≈ 0.139 per minute.
Going from 40 to 10 is a factor of 4, which is two halvings. The time is 2 × 5 = 10 minutes.
No log table needed. Spotting halvings and quarterings saves time in JEE Advanced numericals.
Worked example 2: the average-temperature method
The shortcut treats the rate of cooling as constant over the interval, equal to its value at the average temperature:
(drop in temperature)/(time) = K × (average temperature − Ts)
First interval: 30/5 = 6 °C per minute, and the average temperature is 65 °C. So 6 = K(65 − 20), giving K = 2/15 per minute.
Second interval: the average is 45 °C, so 30/t = (2/15)(45 − 20) = 10/3, giving t = 9 minutes.
The two answers differ by 10 percent. The shortcut is weak here because the excess falls from 40 to 10 within one interval, so the rate changes a lot.
Now try a gentle case: a body cools from 61 °C to 59 °C in 1 minute with surroundings at 20 °C. How long from 51 °C to 49 °C? The average method gives 2/1 = K × 40, so K = 0.05, then 2/t = 0.05 × 30 and t ≈ 1.333 minutes. The exact law gives ln(31/29)/ln(41/39) ≈ 1.334 minutes. When the drop is small compared with the excess, the two agree.
Which mistakes cost marks?
Halving the temperature instead of the excess. From 80 °C to 50 °C is not "nearly half". The excess went from 60 to 30, which is exactly half.
Forgetting the surroundings drift. The NCERT lab manual lists a changing enclosure temperature as a source of error, which is why it asks you to note the enclosure temperature at the start and the end.
Mixing ln and log₁₀. If you plot log₁₀, the slope is −k/2.303, not −k.
Using the law for very hot bodies. A red-hot iron rod is far outside the small-difference range, so expect Stefan's law instead.
Ignoring heat capacity. The lab manual writes the rate of fall of temperature as (k/ms)(θ − θ₀), so two bodies with the same surface but different mass or specific heat cool at different rates. For mixing problems, see our guide to calorimetry and heat transfer.
A practice routine
Redo NCERT exercise 10.20 both ways from memory, then write one sentence on why the answers differ.
Solve three "find the temperature" and three "find the time" problems. Before each, predict whether the two methods will agree.
Derive T − Ts = (T₀ − Ts)e^(−kt) from the differential equation without notes. Our post on differential equations for JEE covers separating variables.
If you do the CBSE experiment, plot both the cooling curve and the log graph, and find k from the slope.
If a step in a worked solution does not make sense, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer, so you can see why each line follows.
Tonight, solve NCERT exercise 10.20 both ways and check that you get 10 and 9 minutes.
