JEE Main · 2012 · 29 Apr 2012 · Offline · Heat & Thermodynamics

Newton's law of cooling says the excess of temperature over the surroundings falls at a rate proportional to itself, $\frac{d\theta }{dt}=-k(\theta -{\theta }_0)$. Separating and integrating gives $\ln (\theta -{\theta }_0)=-kt+\text{constant}$. So a plot of ${\log }_e(\theta -{\theta }_0)$ against $t$ is a straight line that begins at a positive intercept and slopes steadily downward with gradient $-k$.