Question

A cup of coffee (temperature = 190°F) is placed in a room whose temperature is 70°F. After five minutes, the temperature of the coffee has dropped to 160°F. How many more minutes must elapse before the temperature of the coffee is 130°F?

Answer 1

As Newton's law of cooling is $T(t)=T_e+(T_0-T_e)e^{-kt}$ where $T_e$ is a temperature of environment, $T_0$ is a temperature at time $t=0$ . When

$k=-\frac{1}{t}\ln\frac{T_1-T_e}{T_0-T_e}$ then $t_2=t_1(1-\ln{\frac{T_1-T_e}{T_2-T_e}}/{\ln\frac{T_1-T_e}{T_0-T_e}})$ where $T_0=190^0F,T_1=160^0F,T_2=130^0F, t_1=5min$ , then $t_2\approx12min$