Answer to Question #290004 in Differential Equations for aisme

Question

A thermometer is taken from a room where the temperature is 20°C to the outdoors,

where the temperature is 5°C. After one minute the thermometer read 12°C.Use

Newton’s Law of Cooling to answer the following questions:

a) What will the reading on the thermometer be after one more minute?

b) When will the thermometer read 6°C?

Answer 1

As per Newton's Law of Cooling, the rate of the cooling is always proportional to the temperature difference between the object and its surroundings.

$\frac{dT}{dt}=-k(T_0-T_s)$

$T_t=T_s+(T_0-T_s)e^{-kt}$

$T_0=20°C$ And $T_s=5°C$

At t=60, $T_{60}=12°C$

$12=5+(20-5)e^{-60k}$

$12-5=15e^{-60k}$

$\frac{7}{15}=e^{-60k}$

$k=-\frac{ln(\frac{7}{15})}{60}$

$k=0.0127023342$

a)

After another one minute, $\therefore t=120s$

$T_t=T_s+(T_0-T_s)e^{-kt}$

$T_{120}=5+(15)e^{-0.0127023342*120}$

$=5+0.01611786424$

$=5.01611786424°C$

b)

When temperature will read 6°C

$T_t=T_s+(T_0-T_s)e^{-kt}$

$6=5+(15)e^{-0.0127023342t}$

$1=(15)e^{-0.0127023342t}$

$e^{-0.0127023342t}=\frac{1}{15}=0.06666666667$

$t=-\frac{ln 0.06666666667}{0.0127023342}$

t=213.19s

=3.55minutes