Newton's law of cooling

Separation.
$$\frac{dy}{70-y} = k\,dt \implies -\ln|70-y| = kt + C \implies y(t) = 70 - Ae^{-kt}.$$

Initial condition. $y(0)=40$ gives $40 = 70 - A$, so $A=30$.

Finding $k$. $y(10)=48$ gives
$$48 = 70 - 30e^{-10k} \implies e^{-10k} = \frac{22}{30} = \frac{11}{15} \implies k = -\frac{1}{10}\ln\frac{11}{15} \approx 0.0310.$$

At $t=25$.
$$y(25) = 70 - 30e^{-25k} \approx 70 - 30e^{-0.775} \approx 70 - 13.82 = 56.18^\circ\text{F}.$$