Newton's Law of Cooling Formula: dT/dt = −k(T − Ts)

Newton's law of cooling formula is $\frac{dT}{dt} = -k(T - T_s)$ : the rate at which a hot body loses temperature is proportional to how much hotter it is than its surroundings. A cup of tea at 90 °C in a 25 °C room cools fast at first, then more and more slowly as the gap shrinks.

Solving that rate equation gives the exponential cooling curve $T = T_s + (T_0 - T_s)e^{-kt}$ , which the calculator above uses. Class 11 students also use an average form that needs no logarithms. The law works well when the temperature difference is small and heat is lost mainly by convection.

Quick facts

Rate form
$\frac{dT}{dt} = -k(T - T_s)$
Solution
$T = T_s + (T_0 - T_s)e^{-kt}$
Class 11 average form
$\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_s\right)$
Unit of k
s⁻¹ (or min⁻¹), dimensions [T⁻¹]
Valid when
Excess temperature is small, surroundings steady
Graph
ln(T − Tₛ) against t is a straight line of slope −k

Terms in Newton's law of cooling formula

What each symbol means
Symbol	Meaning	Unit
$\frac{dT}{dt}$	How fast the temperature changes (negative while cooling)	°C/s or K/s
$k$	Cooling constant, set by surface area, material and air flow	s⁻¹
$T$	Body temperature at time t	°C or K
$T_s$	Steady temperature of the surroundings	°C or K
$T_0$	Body temperature at t = 0	°C or K

Only differences of temperature appear, so you can work in °C or K without converting, as long as you are consistent. The constant k is larger for a body with more surface area, a thinner wall or a breeze blowing over it.

The Class 11 average method

Board questions often give two cooling intervals and ask for a time. Replace the instantaneous rate by an average rate, and the instantaneous temperature by the average of the start and end values:

$\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_s\right)$

Example 1: two intervals

A body cools from 80 °C to 60 °C in 5 minutes in a room at 20 °C. How long will it take to cool from 60 °C to 40 °C?

First interval: 20/5 = k(70 − 20), so 4 = 50k and k = 0.08 per minute.

Second interval: 20/t = 0.08 × (50 − 20) = 2.4.

t = 20 / 2.4 ≈ 8.3 minutes.

The second drop of 20 °C takes longer because the body is closer to room temperature.

Exponential cooling examples

Example 2: temperature after a set time

Tea at 90 °C is left in a 25 °C room. Its cooling constant is 1.0 × 10⁻³ s⁻¹. Find its temperature after 5 minutes.

t = 300 s, so kt = 0.30 and e⁻⁰·³ ≈ 0.741.

$T = 25 + (90 - 25)(0.741) = 25 + 48.2 \approx 73\ ^\circ\text{C}$

Example 3: finding k from data

Coffee cools from 85 °C to 55 °C in 600 s in a 20 °C room. Find k.

$k = \frac{1}{t}\ln\frac{T_0 - T_s}{T - T_s} = \frac{1}{600}\ln\frac{65}{35} = \frac{0.619}{600} \approx 1.0\times10^{-3}\ \text{s}^{-1}$

Reading the cooling graph

A plot of T against t is a curve that drops steeply at first and flattens toward $T_s$ , never quite reaching it. Take logarithms of the solution and you get $\ln(T - T_s) = \ln(T_0 - T_s) - kt$ . A plot of $\ln(T - T_s)$ against t is therefore a straight line of slope −k. Lab experiments use this to test the law and measure k.

The time to fall halfway to room temperature is the same at every stage: ln 2 / k. For k = 1.0 × 10⁻³ s⁻¹ that is about 690 s.

Where the law comes from and its limits

A body loses heat at a rate $\frac{dQ}{dt} = -hA(T - T_s)$ , where h is a heat transfer coefficient and A the surface area. Since $Q = mcT$ for a body of mass m and specific heat c, dividing by mc gives the law with $k = \frac{hA}{mc}$ . The link between heat and temperature change is covered on the heat formula page.

Radiation follows a fourth power law, so Newton's law is only an approximation for it, good when T and Tₛ are close. Conduction through a solid wall is handled instead by the thermal conductivity relation.

Common mistakes

Using the body temperature alone instead of the difference T − Tₛ.
Mixing minutes and seconds between k and t; the product kt must have no unit.
Using log base 10 where the formula needs the natural log.
Applying the law to very large temperature differences.

More heat and thermodynamics formulas are collected on the thermodynamics formula sheet. The unit of k matches the dimensions of frequency, [T⁻¹].

Frequently asked questions

What is Newton's law of cooling formula in exponential form?

T = Tₛ + (T₀ − Tₛ)e^(−kt). It comes from integrating dT/dt = −k(T − Tₛ).

What is Newton's law of cooling formula for Class 11?

The average form (T₁ − T₂)/t = k[(T₁ + T₂)/2 − Tₛ] is used for two cooling intervals without logarithms.

What is the unit of the cooling constant k?

Per second (s⁻¹), or per minute if time is in minutes. Its dimensional formula is [T⁻¹].

Does a body ever reach room temperature by Newton's law?

In the ideal law the difference only approaches zero. In practice it becomes too small to measure after a few multiples of 1/k.

When does Newton's law of cooling fail?

It is poor for large temperature differences, where radiation dominates, and when the surroundings themselves heat up.