Question

Can Newton's Law of Cooling be used to describe heating?

Answer 1

Hint: Newton’s law of cooling states that the cooling rate of a hot body is proportional to the temperature difference between the body and its surrounding. Newton’s law of cooling describes how an exposed body changes its temperature through radiation with time.

Complete step by step answer:
Let ${T_0}$ is the initial temperature of a given body and ${T_S}$ is the temperature of the surrounding.
Apply the Newton’s law of cooling, the temperature of the give body at a time t is
$T\left( t \right) = {T_S} + \left( {{T_0} - {T_S}} \right){e^{ - kt}}$
Where, $k$ is a constant.
When ${T_0} > {T_S}$, the 2nd term of the above formula decreases with time. So, this describes cooling of the given body.
When ${T_0} > {T_S}$, the 2nd term of the above formula decreases with time. So, this describes cooling of the given body.
When ${T_0} < {T_S}$, the 2nd term of the above formula increases with time. So, this describes the heating of the given body.
Hence, Newton's Law of Cooling can be used to describe heating of a given body.

Additional Information:
The above formula can be derived as follows.
According to the Newton’s cooling law, the rate of cooling $\dfrac{{dT}}{{dt}} \propto \left( {{T_0} - {T_S}} \right)$
$ \Rightarrow \dfrac{{dT}}{{dt}} = - k\left( {{T_0} - {T_S}} \right)$
Where, $k$ is a constant.
Now rearrange the above equation to apply the integration.
$ \Rightarrow \dfrac{{dT}}{{\left( {T - {T_S}} \right)}} = - kdt$
Now take integration on both sides of the above equation.
Initial time $t = 0$ and initial temperature $T = {T_0}$
Final time $t = t$ and final temperature $T = T\left( t \right)$
$ \Rightarrow \int_{{T_0}}^{T\left( t \right)} {\dfrac{{dT}}{{T - {T_S}}}} = \int_0^t { - kdt} $
Simplify the above integration.
\[ \Rightarrow \left[ {\log \left( {T - {T_S}} \right)} \right]_{{T_0}}^{T\left( t \right)} = \left[ { - kt} \right]_0^t\]
Further calculate the above equation.
$ \Rightarrow \log \left( {T\left( t \right) - {T_S}} \right) - \log \left( {{T_0} - {T_S}} \right) = - kt - 0$
$ \Rightarrow \log \left( {\dfrac{{T\left( t \right) - {T_S}}}{{{T_0} - {T_S}}}} \right) = - kt$
Now take exponent on the both side of the above equation and remember that ${e^{\log a}} = a$
$ \Rightarrow \dfrac{{T\left( t \right) - {T_S}}}{{{T_0} - {T_S}}} = {e^{ - kt}}$
Multiply $\left( {{T_0} - {T_S}} \right)$ on both sides of the above equation and then add ${T_S}$ on both sides.
$ \Rightarrow T\left( t \right) = {T_S} + \left( {{T_0} - {T_S}} \right){e^{ - kt}}$

Note: Note that the temperature difference between the body and its surroundings must be small. Also, the surrounding temperature must be constant during the entire process of cooling of the body.