What is Newton’s Law of Cooling?

Newton's Law of Cooling states that how fast an object cools over a period is proportional to the difference between the temperature of the object and its surroundings.

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Another differential equation (DE) application is Newton's Law of Cooling. Sir Isaac Newton was interested in quantitatively finding the heat loss in an object; hence, he derived a model that would represent this event.

Modeling Newton's Law of Cooling

Consider an object in space like a cup of hot water in a cold room. From thermodynamics, we know that this hot water will eventually cool off. How fast this water cools over a period $\frac{d T}{d t}$ is proportional to the difference between the temperature of the object $T_{o}$ and its surroundings $T_{s}$ . Representing it using equations, we have:

$\frac{d T}{d t} \propto T_{o} - T_{s}$
$\frac{d T}{d t} = k (T_{o} - T_{s})$ , where $k$ is the constant of proportionality

We can solve this DE using the separation of variables technique and expressing the solution in its exponential form:

$\frac{d T}{d t} = k (T_{o} - T_{s})$
$\frac{d T}{T_{o} - T_{s}} = k \cdot d t$
$\ln (T_{o} - T_{s}) = k t + \ln C$
$\ln (T_{o} - T_{s}) - \ln C = k t$
$\ln \frac{T_{o} - T_{s}}{C} = k t$
$e^{\ln \frac{T_{o} - T_{s}}{C}} = e^{k t}$
$\frac{T_{o} - T_{s}}{C} = e^{k t}$
$T_{o} = C e^{k t} + T_{s}$

This equation, $T_{o} = C e^{k t} + T_{s}$ , is Newton's Law of Cooling. It consists of the following constants and variables:

$C$ is the initial value
$k$ is the constant of proportionality
$t$ is time
$T_{o}$ is temperature of the object at time $t$
$T_{s}$ is the constant temperature of the surrounding environment.

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Example

Let's say we have an object in a room constantly at 15.6ºC. When we place it inside at there first, its initial temperature is 37.8ºC. After 10 mins, it cooled off to 32.2ºC. How many more minutes will it take to be at 26.7ºC?

Identify

In this problem, we need to determine the number of minutes it will take to reach 26.7ºC if 10 minutes have passed.

To start, let's list the important details:

Initial Condition: $T_{1} = 37.8$ when $t_{1} = 0$
Secondary Condition: $T_{2} = 32.2$ when $t_{2} = 10$
Unknown Condition: $T_{3} = 26.7$ when $t_{3}$
Room Temperature: $T_{S} = 15.6$

Find the Model

The model that we will use is Newton's Law of Cooling: $T_{o} = C e^{k t} + T_{s}$ , which we derived its general solution earlier.

After identifying this model (general solution), we need to find the constants $C$ and $k$ based on the conditions. Another way of saying this is we need to find a particular solution based on the given criteria:

For $C$ , let's use the initial criteria:

$T_{o} = C e^{k t} + T_{s}$
$37.8 = C e^{k (0)} + 15.6$
$C = 22.2$

For $k$ , consider the secondary condition:

$T_{o} = C e^{k t} + T_{s}$
$32.2 = 22.2 e^{10 k} + 15.6$
$k = - 0.02907$

The constant of proportionality is negative; hence, we can confirm the object is cooling off.

Now that we know these constants, we can now form the model

$T_{o} = 22.2 e^{- 0.02907 t} + 15.6$

This expression is the function we use to determine the object's temperature at any given time. We can see a graphical view of the function below.

Solution Curve

Evaluate

To answer the problem, let's now consider the unknown condition. We will substitute $T_{o} = 26.7$ to the equation and solve for the remaining variable $t$ :

$T_{o} = 22.2 e^{- 0.02907 t} + 15.6$
$26.7 = 22.2 e^{- 0.02907 t} + 15.6$
$t = 23.8$

The time it takes for the object to reach 26.7ºC is 23.8 mins.

How many more minutes will it take after 10 minutes have passed? It would be 23.8-10 = 13.8 minutes.

Summary

How fast an object cools over a period $\frac{d T}{d t}$ is proportional to the difference between the temperature of the object $T_{o}$ and its surroundings $T_{s}$ . Representing it using equations, we have: $\frac{d T}{d t} = k (T_{o} - T_{s})$ , where $k$ is the constant of proportionality

The general equation, $T_{o} = C e^{k t} + T_{s}$ , is Newton's Law of Cooling. It consists of the following constants and variables: $C$ is the initial value, $k$ is the constant of proportionality, $t$ is time, $T_{o}$ is the temperature of the object at time $t$ , and $T_{s}$ is the constant temperature of the surrounding environment.

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Created On

June 5, 2023

Updated On

February 23, 2024

Contributors

Edgar Christian Dirige

Founder

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