JEE Main & Advanced Physics Transmission of Heat Newton's Law of Cooling

When the temperature difference between the body and its surrounding is not very large i.e. \[T-{{T}_{0}}=\Delta T\]then \[{{T}^{4}}-T_{0}^{4}\] may be approximated as \[4T_{0}^{3}\Delta T\]

By Stefan?s law, \[\frac{dT}{dt}=\frac{A\varepsilon \sigma }{mc}[{{T}^{4}}-T_{0}^{4}]\]

Hence \[\frac{dT}{dt}=\frac{A\varepsilon \sigma }{mc}4T_{0}^{3}\Delta T\]\[\Rightarrow \]\[\frac{dT}{dt}\propto \Delta T\] or  \[\frac{d\theta }{dt}\propto \theta -{{\theta }_{0}}\]

i.e., if the temperature of body is not very different from surrounding, rate of cooling is proportional to temperature difference between the body and its surrounding. This law is called Newton's law of cooling.

(1) Greater the temperature difference between body and its surrounding greater will be the rate of cooling.

(2) If \[\theta ={{\theta }_{0}}\], \[\frac{d\theta }{dt}=0\] i.e. a body can never be cooled to a temperature lesser than its surrounding by radiation.

(3) If a body cools by radiation from \[\theta _{1}^{o}C\] to \[\theta _{2}^{o}C\] in time t, then \[\frac{d\theta }{dt}=\frac{{{\theta }_{1}}-{{\theta }_{2}}}{t}\] and \[\theta ={{\theta }_{av}}=\frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}\]. The Newton's law of cooling becomes \[\left[ \frac{{{\theta }_{1}}-{{\theta }_{2}}}{t} \right]=K\left[ \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-{{\theta }_{0}} \right]\].

This form of law helps in solving numericals.

(4) Practical examples

(i) Hot water loses heat in smaller duration as compared to moderate warm water.

(ii) Adding milk in hot tea reduces the rate of cooling.