Variable Separable

Newton's law of cooling states that the rate of temperature change is proportional to the difference between the object's temperature and the surrounding temperature T_s, giving a separable first-order equation. The key idea is to separate the temperature difference from time, treating T - T_s as the relevant variable. Let u = T - T_s, so \frac{du}{dt} = \frac{dT}{dt} = -ku. This separates to \frac{du}{u} = -k,dt, and integrating gives \ln|u| = -kt + C_1, hence u = Ce^{-kt}. Restoring u = T - T_s gives T - T_s = Ce^{-kt}. Applying the initial temperature T_0 at t = 0 gives C = T_0 - T_s, so T = T_s + (T_0 - T_s)e^{-kt}. Option with e^{kt} would describe heating without bound, contradicting cooling. Option T = T_0 e^{-kt} ignores the ambient temperature entirely. Option with a minus sign before (T_0 - T_s) gives the wrong initial value at t = 0. This is the standard JEE Advanced cooling application. As a final plausibility check, as t \to \infty the exponential vanishes and T \to T_s, correctly modeling that the coffee eventually reaches room temperature.