# Newton’s Law of Cooling

## Solved Problems

### 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.

Solution.

First of all, we should emphasize the difference with the case of the constant environment temperature. In the last case, a body's temperature will be approaching the environment temperature an infinitely long time. In the given problem, the environment temperature increases linearly, and therefore, sooner or later both the temperatures become equal in some time. Thus, the problem has a solution.

The cooling process is described by the differential equation:

\[\frac{{dT}}{{dt}} = k\left( {{T_S} - T} \right).\]

In the given case \({T_S} = {T_{S0}} + \beta t.\) Hence, the last equation can be written in the form:

\[\frac{{dT}}{{dt}} = k\left( {{T_{S0}} + \beta t - T} \right)\;\; \text{or}\;\; T' + kT = k{T_{S0}} + k\beta t.\]

We have obtained a linear differential equation, which can be solved using, for example, the integrating factor:

\[u\left( t \right) = {e^{\int {kdt} }} = {e^{kt}}.\]

The general solution of the equation is written as

\[T\left( t \right) = \frac{{\int {{e^{kt}}\left( {k{T_{S0}} + k\beta t} \right)dt} + C}}{{{e^{kt}}}} = \frac{{k{T_{S0}}\int {{e^{kt}}dt} + k\beta \int {{e^{kt}}tdt} + C}}{{{e^{kt}}}}\]

The second integral in the numerator can be calculated by integrating by parts:

\[\int {\underbrace {{e^{kt}}}_{u'}\underbrace t_vdt} = \left[ {\begin{array}{*{20}{l}}
{u' = {e^{kt}}}\\
{u = \frac{1}{k}{e^{kt}}}\\
{v = t}\\
{v' = 1}
\end{array}} \right]
= \frac{1}{k}{e^{kt}}t - \int {\frac{1}{k}{e^{kt}}dt}
= \frac{1}{k}{e^{kt}}t - \frac{1}{{{k^2}}}{e^{kt}}
= \frac{1}{k}{e^{kt}}\left( {t - \frac{1}{k}} \right).\]

So we get the cooling law of the body in the following form:

\[T\left( t \right) = {T_{S0}} + \beta t - \frac{\beta }{k} + C{e^{ - kt}}.\]

The constant \(C\) can be found from the initial condition \(T\left( {t = 0} \right) = {T_0}.\) Then

\[C = {T_0} - {T_{S0}} + \frac{\beta }{k}.\]

Hence, the body's cooling law is given by the formula

\[T\left( t \right) = {T_{S0}} + \beta t - \frac{\beta }{k} + \left( {{T_0} - {T_{S0}} + \frac{\beta }{k}} \right){e^{ - kt}}.\]

At a certain moment \(\tau,\) the temperature of the body and temperature of the surrounding environment will become equal:

\[T\left( \tau \right) = {T_{S0}} + \beta \tau .\]

Thus, the time \(\tau\) is found from the equation:

\[{\cancel{{T_{S0}} + \beta \tau} = \cancel{{T_{S0}} + \beta \tau} - \frac{\beta }{k} + \left( {{T_0} - {T_{S0}} + \frac{\beta }{k}} \right){e^{ - k\tau }},\;\;} \Rightarrow
\left( {{T_0} - {T_{S0}} + \frac{\beta }{k}} \right){e^{ - k\tau }} = \frac{\beta }{k},\;\; \Rightarrow
\frac{k}{\beta }\left( {{T_0} - {T_{S0}} + \frac{\beta }{k}} \right) = {e^{k\tau }},\;\; \Rightarrow
\frac{k}{\beta }\left( {{T_0} - {T_{S0}}} \right) + 1 = {e^{k\tau }},\;\; \Rightarrow
\tau = \frac{1}{k}\ln \left[ {\frac{k}{\beta }\left( {{T_0} - {T_{S0}}} \right) + 1} \right].\]

We can estimate the time \(\tau\) for some typical values of the parameters:

\[{T_{S0}} = 20^{\circ}C,\;\;\; \kern-0.3pt k = \frac{1}{5}\,\text{min}^{-1},\;\;\; \kern-0.3pt
\beta = 2\,\frac{\text{degrees}}{\text{min}},\;\;\; {T_0} = 200^{\circ}C.\]

As a result, we have:

\[\tau = \frac{1}{k}\ln \left[ {\frac{k}{\beta }\left( {{T_0} - {T_{S0}}} \right) + 1} \right]
= \frac{1}{{\frac{1}{5}}}\ln \left[ {\frac{{\frac{1}{5}}}{2}\left( {200 - 20} \right) + 1} \right]
= 5\ln \left[ {\frac{1}{{10}} \cdot 180 + 1} \right] = 5\ln 19 \approx 5 \cdot 2.944 \approx 14.77\left[ {\text{min}} \right].\]