02 - Time for the thermometer reading to drop to within half degree of the air temperature

$\dfrac{dT}{dt} = -k(T - 20)$

$\dfrac{dT}{T - 20} = -k\,dt$

$\ln (T - 20) = -kt + \ln C$

$\ln (T - 20) = \ln e^{-kt} + \ln C$

$\ln (T - 20) = \ln Ce^{-kt}$

$T - 20 = Ce^{-kt}$

$T = 20 + Ce^{-kt}$
 

When t = 0, T = 75°
$75 = 20 + Ce^{0}$

$C = 55$
 

Hence,
$T = 20 + 55e^{-kt}$
 

When t = 4, T = 30°
$30 = 20 + 55e^{-4k}$

$\frac{10}{55} = e^{-4k}$

$e^{-k} = \left( \frac{2}{11} \right)^{1/4}$
 

Thus,
$T = 20 + 55\left( \frac{2}{11} \right)^{t/4}$
 

When T = 20.5°F
$20.5 = 20 + 55\left( \frac{2}{11} \right)^{t/4}$

$\dfrac{0.5}{55} = \left( \frac{2}{11} \right)^{t/4}$

$\ln \dfrac{0.5}{55} = \ln \left( \frac{2}{11} \right)^{t/4}$

$\ln (\frac{1}{110}) = \frac{1}{4}t \, \ln (\frac{2}{11})$

$t = \dfrac{4\ln (\frac{1}{110})}{\ln (\frac{2}{11})}$

$t = 11.029 ~ \text{minutes}$           answer