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.