According to Newton’s Law of cooling, the time rate of change of temperature is proportional to the temperature difference.

$\dfrac{dT}{dt} = -k(T - T_s)$

$\dfrac{dT}{dt} = -k(T - 10)$

$\dfrac{dT}{T - 10} = -k \, dt$

$\ln (T - 10) = -kt + \ln C$

$\ln (T - 10) = \ln e^{-kt} + \ln C$

$\ln (T - 10) = \ln Ce^{-kt}$

$T - 10 = Ce^{-kt}$

$T = 10 + Ce^{-kt}$

When t = 0, T = 70°F

$70 = 10 + C$

$C = 60$

Hence,

$T = 10 + 60e^{-kt}$

When t = 3 min, T = 25°F

$25 = 10 + 60e^{-3k}$

$15 = 60e^{-3k}$

$\frac{15}{60} = e^{-3k}$

$e^{-k} = (\frac{1}{4})^{1/3}$

Thus,

$T = 10 + 60(\frac{1}{4})^{t/3}$

After 6 minutes, t = 6

$T = 10 + 60(\frac{1}{4})^2$

$T = 13.75^\circ F$ *answer*