## Definition of Newton's Law of Cooling

Newton's law of cooling explains the rate at which an object loses heat to its environment. It connects the temperature difference between the object and the environment with the rate of heat loss by the object.

**Newton's law of cooling** states that the rate of heat loss of a body is directly proportional to the difference in the temperature between the body and its surroundings.

Put simply, Newton's law of cooling describes how fast an object cools down given the temperature difference with its environment: the bigger the temperature difference, the faster our object cools down. For example, it can help us explain how fast a cup of boiling water will cool down in a kitchen at room temperature.

Newton's law of cooling is not always perfectly accurate, there are a few conditions that must hold.

The nature of heat loss must remain the same whilst the heat loss takes place.

The temperature of the surroundings must remain constant whilst the heat loss takes place.

The difference between temperature and the surroundings must be small.

In many situations, these approximations are valid, however, in certain methods of heat transfer, such as convection or radiation, the law is less applicable.

There can be some confusion over the difference between heat and temperature and what this law is actually telling us. In fact, Newton himself first misstated the law, as the difference between heat and temperature was not fully known at the time. So, before moving on, let's recap a couple of definitions.

**Heat**** **is energy that is transferred from one object or source to another. Heat occurs when there is a difference in temperature between two objects. Unlike in day-to-day usage of the word, in physics heat is a process, not a property of a material.

**Temperature** is a measure of the average kinetic energy of atoms or molecules in a material. It is a property of an object itself.

## Example of Newton's Law of Cooling

To gain some intuition on what Newton's law of cooling tells us, consider having a cup of soup. We all know that the soup will be too hot to eat for the first minute or so, but eventually, it reaches a temperature we can enjoy it at. We then can eat the soup leisurely over say ten minutes, and it stays warm throughout. Does this match up with what Newton's Law tells us? Well, initially, the soup is much hotter than the air around it and from Newton's Law, we know that the rate of temperature loss will be highest when the temperature difference is highest. So, we expect the rate of cooling to be quickest when the soup is hottest, which is precisely what we see in reality. Newton's law says that as the temperature of the soup cools toward room temperature, its rate of cooling slows. This is why it stays at a nice warm temperature for much longer than it does at its hottest temperature. Lucky us!

## Formula for Newton's Law of Cooling

Now we understand the basic ideas of Newton's law of cooling, let's dive into what it says mathematically! Written as a equation, Newton's law states the following: \[\frac{\text{d}T}{\text{d}t}=-r(T-T_\text{env}),\] where:

\(\frac{\text{d}T}{\text{d}t}\) is the rate of change in temperature of the object (units: \(\frac{\mathrm{K}}{\mathrm{s}}\)),

\(T\) is the temperature of the object (units: \(\mathrm{K}\)),

\(T_\text{env}\) is the temperature of the environment (units: \(\mathrm{K}\)),

\(r\) is a constant dependent on the environmental material, the object's material, the object's heat capacity, and the object's surface area, called the coefficient of heat transfer (unit: \(\frac{1}{\mathrm{s}}\)).

This equation is a type of differential equation and can be solved to find the temperature of the body \(T(t)\) as a function of time \(t\). However, solving differential equations is not something you will do in high school, and in this article, we will just state its solution:

\[T(t)=T_\text{env}+(T_0-T_\text{env})\text{e}^{-rt},\]

where \(T_0\) is the initial temperature of the object and \(\text{e}\) is Euler's constant.

It is this second equation that we will focus on, as it allows us to directly calculate temperatures and plot them as a graph over time. Let's have a look at some of the general properties of this equation.

Straight away, we see that setting \(t=0\,\mathrm{s}\) means the exponential term is just one and we get\[T(0\,\mathrm{s})=T_\text{env}+(T_0-T_{env})\cdot 1=T_0.\]

On the other hand, as \(t\to\infty\), we get that \(T\to T_\text{env}\) as we would expect as two objects reach the same temperature or

**thermal equilibrium.**Objects with a higher coefficient of heat transfer lose heat faster than those with a lower coefficient. The larger the surface area of an object, the faster it will cool down as the exponent becomes more negative. This is precisely what happens in reality, as there is more surface area for heat to pass between the body and its environment.

## Calculation with Newton's Law of Cooling

We will now walk through a calculation using the temperature formula introduced above.

Q: Consider a bath of cold water at a temperature of \(T_\text{bath}=280\,\text{K}\). If a spherical hot iron ball at an initial temperature of \(T_0=350\,\text{K}\) is dropped into the bath, what will the temperature of the ball be 10.0 seconds later? Take the heat transfer coefficient as \(r=0.250\,\frac{1}{s}\) and assume the bath is big enough that the temperature of the bath water does not change.

A: First write down the formula for the temperature from Newton's Law:\[T(t)=T_\text{bath}+(T_0-T_\text{bath})\text{e}^{-rt},\]where now our environment is the bath. The question has given us all the values we need to solve this problem so we can just plug the values into the temperature formula as follows:

\[T(10\,\mathrm{s})=280\,\text{K}+(350\,\text{K}-280\,\text{K} )\text{e}^{-0.250\,\frac{1}{\mathrm{s}}\cdot 10.0\,\text{s}}=286\,\mathrm{K}.\]

As we see, after \(10\,\mathrm{s}\), the object has almost cooled down to the temperature of the environment.

## Graph of Newton's Law of Cooling

To get an overall picture of how an object following Newton's law cools down over time, we can plot a graph. The temperature-time graph for an object following Newton's law of cooling shows an exponential decay. Notice below how the graph begins with a very large gradient, before slowly curving to have a much smaller gradient until it reaches a temperature very close to its surroundings. The gradient of the graph represents the rate of change in temperature

and is given by the differential equation formula of Newton's law of cooling,\[\frac{\text{d}T}{\text{d}t}=r(T-T_\text{env}).\]As stated by the law, the gradient is initially very high, but decreases as the temperature difference decreases. This is seen in the graph as it begins much steeper, but flattens out as the temperature drops. This happens because the object begins by losing heat to the surrounding environment rapidly, but as it cools down, the rate at which heat enters the surrounding environment slows down. Once the graph levels off, the object's temperature is very close to the surrounding temperature.

In the graph above, data from the previous example calculation has been plotted. It starts at the initial temperature of \(350\,\text{K}\) and tends towards the temperature of the bath \(280\,\text{K}\). In the graph below, the rate of temperature change has been plotted by plotting the gradient of the upper graph. Note it begins extremely negative and levels out towards zero from below as expected.

## Newton's Law of Cooling - Key takeaways

- Newton's law of cooling governs objects cooling down due to heat transfer to a cooler environment, such as a mug of coffee in a room cooling down to room temperature.
- The formula associated with Newton's law of cooling is \[T(t)=T_\text{env}+(T_0-T_\text{env})\text{e}^{-rt},\] where \(T(t)\) is the temperature of the object at a time \(t\), \(T_\text{env}\) is the temperature of the environment, and \(T_0\) is the initial temperature of the object.
- The differential equation that says the same but in another form (that is sometimes more useful) states\[\frac{\text{d}T}{\text{d}t}=-r(T-T_\text{env}).\]
- The rate of cooling is also determined by the object's surface area, heat capacity, and the materials involved, and this information is all found in the coefficient of heat transfer \(r\).
- The graph representing Newton's law of cooling is a negative exponential. This means it begins with a large negative gradient, before becoming flatter as the temperature decreases and the object approaches thermal equilibrium with its surroundings.

## References

- Photo by Hanna Balan on Unsplash

###### Learn with 9 Newton's Law of Cooling flashcards in the free StudySmarter app

We have **14,000 flashcards** about Dynamic Landscapes.

Already have an account? Log in

##### Frequently Asked Questions about Newton's Law of Cooling

What is Newton's law of cooling?

Newton's law of cooling is a law that governs how fast an object cools down to the temperature of its surroundings. It states that the rate of heat loss is directly proportional to the temperature difference between an object and its surroundings.

What is Newton's law of cooling used for?

Newton's law of cooling is used for calculating the time it takes for an object to reach a temperature below its current temperature. For example, it can be used for calculating time of deaths in forensic investigations, by comparing a body's current temperature to its probable temperature before death

What is Newton's law of cooling formula?

The formula of Newton's law of cooling is

*T*(*t*) = *T*_{s} + (*T*_{0} - *T*_{s})e^{-rt}*,*

where:

*T*(*t*) is the temperature of an object at a time *t,*

*T*_{s} is the temperature of the surrounding environment,

*T*_{0} is the initial temperature of the object, and

*k* is a constant of proportionality determined by the surface area of the object and its heat transfer coefficient

What is *k* in Newton's law of cooling?

In Newton's law of cooling, *k* is a constant of proportionality known as the coefficient of heat transfer defined by the surface area of the object cooling down and the materials of the object and environment.

How to solve Newton's law of cooling?

Newton's law of cooling can be solved in its differential form d*T*/d*t* = -*r* ( *T* - *T*_{env} ) by separating the variables and then integrating. This then gives a formula for the temperature:

*T*(*t*) = *T*_{s} + (*T*_{0} - *T*_{s})e^{-rt}

##### About StudySmarter

StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.

Learn more