The rate at which a substance cools in moving air, is proportional to the difference between the temperature of the substance and that of air. The temperature of air is 290 K and the substance cools from 370 K to 330 K in 10 minutes. Then the time to cool the substance upto 295 K is

Newton's Law of Cooling gives the rate of temperature change proportional to the difference from ambient temperature $T_a = 290\mathrm{K}$: $\frac{dT}{dt} = -k(T - 290)$ Separating variables and integrating yields the general solution: $\int \frac{dT}{T - 290} = -k\int dt$ $\ln|T - 290| = -kt + C$ $T(t) = 290 + Ae^{-kt}$ The initial condition $T(0) = 370$ determines the constant $A = 80$, giving the temperature function: $T(t) = 290 + 80e^{-kt}$ Using the condition that $T$ decreases from $370\mathrm{K}$ to $330\mathrm{K}$ in 10 minutes: $330 = 290 + 80e^{-10k}$ $e^{-10k} = \frac{1}{2}$ To find when $T = 295\mathrm{K}$: $295 = 290 + 80e^{-kt}$ $e^{-kt} = \frac{1}{16}$ Noting that $\frac{1}{16} = \left(\frac{1}{2}\right)^4 = (e^{-10k})^4 = e^{-40k}$, we equate exponents: $-kt = -40k$ $t = 40$ The substance cools to $295\mathrm{K}$ after 40 minutes.