A hot liquid contained in a vessel loses its temperature; partly by radiation and partly by convection. When the temperature difference between the liquid and the surrounding is small, the loss of heat is mainly due to convection.
Newton’s law of cooling states that the rate at which the liquid loses its heat is directly proportional to the temperature difference between the liquid and its surrounding.
Consider a liquid at temperature $θ$ loses a small amount of its heat $dQ$ in small time $dt$. Let the temperature of the surrounding be $θ_0$. Then, according to Newton’s law of cooling, \[\text{Rate of heat loss}∝\text{Temperature difference}\] \[\text{i.e.} \;\;\;-\frac{dQ}{dt}∝(θ-θ_0)\] The negative sign indicates that the amount of heat goes on decreasing with time. \[-\frac{dQ}{dt}=K(θ-θ_0)\text{ __(1)}\] Where, $K$ is proportionality constant which value depends upon the nature of the liquid and its surface area exposed to atmosphere.
Let the mass of the liquid be $m$ and its specific heat capacity be $S$. If the fall in temperature of the liquid is $dθ$ in time $dt$, then, \[\frac{dQ}{dt}=mS\frac{dθ}{dt}\text{ __(2)}\] From $(1)$ and $(2)$, \[-mS\frac{dθ}{dt}= K(θ-θ_0)\] \[\frac{dθ}{θ-θ_0}=-\frac{K}{mS}\;dt\] Integrating, \[\int \frac{dθ}{θ-θ_0}=-\frac{K}{mS}\int dt\] \[\log_e(θ-θ_0)=-\frac{K}{mS}\;t+C\text{ __(3)}\] Where, $C$ is the integration constant.
Equation $(3)$ can be written as, \[y=-\frac{K}{mS}x+C\] This is the equation of straight line. If graph between $\log_e(θ-θ_0)$ and $t$ is a straight line, then Newton’s law of cooling is verified.
Limitations of Newton’s Law of Cooling
1. It is valid only if the temperature difference between the liquid and the surrounding is small ($30$ to $35°C$).
2. It is valid for the cooling of liquid only.