Answer to Question #346468 in Differential Equations for sami

Question

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. After one-half minute the thermometer reads 50° F. What is the reading of the thermometer at t  1 min? How long will it take for the thermometer to reach 15° F?

Answer 1

Let $T=$the temperature (in $\degree F$ ) recorded by the thermometer at time $t.$

The equation governing $T$ is obtained from Newton’s Law of Cooling:

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

where $k$ is some constant.

We are given that $T(0) = 70$ and $T( 1/ 2 ) = 50.$

$$\dfrac{dT}{T-10}=kdt$$

Intergrate

$$\int \dfrac{dT}{T-10}=\int kdt$$

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

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

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

$$T(0)=C+10=70=>C=60$$

$$T(1/2)=60e^{k(1/2)}+10=50$$

$$e^{k(1/2)}=2/3$$

$$k=2\ln(\dfrac{2}{3})$$

$$T(t)=60(\dfrac{4}{9})^t+10$$

a)

$$T(1)=60(\dfrac{4}{9})^1+10=\dfrac{110}{3}(\degree F)\approx36.67(\degree F)$$

b)

$$60(\dfrac{4}{9})^t+10=15$$

$$(\dfrac{4}{9})^t=\dfrac{1}{12}$$

$$t=\dfrac{\ln(1/12)}{\ln(4/9)}\ min\approx3.06\ min$$