Question

The rate at which a super computer body cools is proportional to the difference between the temperature of the body and that of the surrounding air. If a body in air at 25°C will cool from 100°C to 75°C in one minute, find its temperature at the end of three minutes.

Answer 1

$\int \frac{dU}{U-25}=k\smallint dt+ln$C

$ln(U-25)=kt+lnC$

$U-25=Ce^{kt}$

For t=0, U=100

For t=1, U=75

$100-25=Ce^{0k}$

C=75

$75-25=75e^{1k}$

$e^k=0.67$

k=-0.4

$U(3)-25=75e^{-0.4x3}$

U(3)=47.6 C