Differential Equations

1st Order Equations

Newton’s Law of Cooling

Page 1
Problem 1
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Problem 2

In the late of \(17\)th century British scientist Isaac Newton studied cooling of bodies. Experiments showed that the cooling rate approximately proportional to the difference of temperatures between the heated body and the environment. This fact can be written as the differential relationship:

\[\frac{{dQ}}{{dt}} = \alpha A\left( {{T_S} – T} \right),\]

where \(Q\) is the heat, \(A\) is the surface area of the body through which the heat is transferred, \(T\) is the temperature of the body, \({{T_S}}\) is the temperature of the surrounding environment, \(\alpha\) is the heat transfer coefficient depending on the geometry of the body, state of the surface, heat transfer mode, and other factors.

As \(Q = CT,\) where \(C\) is the heat capacity of the body, we can write:

\[{\frac{{dT}}{{dt}} = \frac{{\alpha A}}{C}\left( {{T_S} – T} \right) }={ k\left( {{T_S} – T} \right).}\]

The given differential equation has the solution in the form:

\[{T\left( t \right) = {T_S} }+{ \left( {{T_0} – {T_S}} \right){e^{ – kt}},}\]

where \({T_0}\) denotes the initial temperature of the body.

Thus, while cooling, the temperature of any body exponentially approaches the temperature of the surrounding environment. The cooling rate depends on the parameter \(k = {\large\frac{{\alpha A}}{C}\normalsize}.\) With increase of the parameter \(k\) (for example, due to increasing the surface area), the cooling occurs faster (see Figure \(1.\))

Cooling curves at two different values of the parameter k

Figure 1.

Solved Problems

Click on problem description to see solution.

 Example 1

The temperature of a body dropped from \(200^\circ\) to \(100^\circ\) for the first hour. Determine how many degrees the body cooled in one hour more if the environment temperature is \(0^\circ?\)

 Example 2

A body at the initial temperature \({T_0}\) is put in a room at the temperature of \({T_{S0}}.\) The body cools according to the Newton’s law with the constant rate \(k.\) The temperature of the room slowly increases by the linear law:

\[{T_S} = {T_{S0}} + \beta t,\]

where \(\beta\) is the known parameter. Determine the time \(\tau,\) when the body’s temperature and the surrounding environment temperature become equal.

Example 1.

The temperature of a body dropped from \(200^\circ\) to \(100^\circ\) for the first hour. Determine how many degrees the body cooled in one hour more if the environment temperature is \(0^\circ?\)

Solution.

First, we solve this problem for an arbitrary environment temperature and then determine the final body’s temperature when the surrounding environment temperature is \(0^\circ.\)

Let the initial temperature of the heated body be \({T_0} = 200^\circ.\) The further temperature dynamics is described by the formula:

\[
{T\left( t \right) }={ {T_S} + \left( {{T_0} – {T_S}} \right){e^{ – kt}} }
= {{T_S} + \left( {200^\circ – {T_S}} \right){e^{ – kt}}.}
\]

At the end of the first hour the body has cooled to \(100^\circ.\) Therefore, we can write the following relationship:

\[
{T\left( {t = 1} \right) = 100^\circ }={ {T_S} + \left( {200^\circ – {T_S}} \right){e^{ – k \cdot 1}},\;\;}\Rightarrow
{{100^\circ = {T_S} }+{ \left( {200^\circ – {T_S}} \right){e^{ – k}}.}}
\]

After \(2\)nd hour the body’s temperature becomes equal to \(X\) degrees:

\[{X = {T_S} }+{ \left( {200^\circ – {T_S}} \right){e^{ – 2k}}.}\]

Thus, we obtain the system of two equations with three unknowns: \({T_S},\) \(k\) and \(X:\)

\[\left\{ \begin{array}{l}
100 = {T_S} + \left( {200 – {T_S}} \right){e^{ – k}}\\
X = {T_S} + \left( {200 – {T_S}} \right){e^{ – 2k}}
\end{array} \right..\]

We cannot determine uniquely the body’s temperature \(X\) after the \(2\)nd hour from this system. However, we can derive the dependence of \(X\) on the environment temperature \({T_S}.\) Express the function \({e^{ – k}}\) from the first equation:

\[{e^{ – k}} = \frac{{100 – {T_S}}}{{200 – {T_S}}}.\]

Hence,

\[{{e^{ – 2k}} = {\left( {{e^{ – k}}} \right)^2} }={ {\left( {\frac{{100 – {T_S}}}{{200 – {T_S}}}} \right)^2}.}\]

Then the dependence \(X\left( {{T_S}} \right)\) has the form:

\[
{X\left( {{T_S}} \right) = {T_S} }+{ \left( {200 – {T_S}} \right){\left( {\frac{{100 – {T_S}}}{{200 – {T_S}}}} \right)^2} }
= {{T_S} + \frac{{{{\left( {100 – {T_S}} \right)}^2}}}{{200 – {T_S}}}.}
\]
Body's temperature in 2 hours depending on the environment temperature

Figure 2.

If, for example, the surrounding environment temperature is zero degrees, the body’s temperature \(X\) in \(2\) hours will be

\[
{X\left( {{T_S} = 0} \right) }={ 0 + \frac{{{{\left( {100 – 0} \right)}^2}}}{{200 – 0}} }
= {\frac{{10000}}{{200}} }={ 50^\circ.}
\]

In the given example the value of \(X\) depends on \({T_S}\) as shown in Figure \(2.\)

Page 1
Problem 1
Page 2
Problem 2