Question

Frank’s automobile engine runs at 100∘C. One day, when the outside temperature is 15∘C, he turns off the ignition and notes that 5 minutes later, the engine has cooled to 70∘C. When will the engine cool to 40∘C? (Use decimal notation. Give your answer to the nearest whole number.)

Answer 1

Answer:

the engine cool to 40$^{o}C$ at 14.07 minutes

Explanation:

Given information

T(5) = 70$^{o}C$

$T_{0}$ = 100$^{o}C$

C = 15$^{o}C$

Newton's law of cooling :

T(t) = C + ($T_{0}$ - C) $e^{-kt}$

where

T(t) = temperature at any given time

C = surrounding temperature

$T_{0}$ = initial temperature of heated object

k = cooling constant

to find the the time when the engine will be cooled down to 40$^{o}C$, we first need to find the cooling constant, k

when t = 5, T(5) = 70$^{o}C$

so,

T(t) = C + ($T_{0}$ - C) $e^{-kt}$

T(5) = 15 + (100 - 15) $e^{-5k}$

70 = 15 + (85) $e^{-5k}$

$e^{-5k}$ = (70 - 15) / 85

-5k = ln (55/85)

k = - ln (55/85) / 5

k = 0.087

thus, we have the eqaution

T(t) = 15 + (85) $e^{-0.087t}$

now we can determine the time when T(t) = 40$^{o}C$

40 = 15 + (85) $e^{-0.087t}$

$e^{-0.087t}$ = (40-15)/85

-0.0087t = ln (25/85)

t = - ln (25/85)/0.087

t = 14.07 minutes