Answer:
the engine cool to 40[tex]^{o}C[/tex] at 14.07 minutes
Explanation:
Given information
T(5) = 70[tex]^{o}C[/tex]
[tex]T_{0}[/tex] = 100[tex]^{o}C[/tex]
C = 15[tex]^{o}C[/tex]
Newton's law of cooling :
T(t) = C + ([tex]T_{0}[/tex] - C) [tex]e^{-kt}[/tex]
where
T(t) = temperature at any given time
C = surrounding temperature
[tex]T_{0}[/tex] = initial temperature of heated object
k = cooling constant
to find the the time when the engine will be cooled down to 40[tex]^{o}C[/tex], we first need to find the cooling constant, k
when t = 5, T(5) = 70[tex]^{o}C[/tex]
so,
T(t) = C + ([tex]T_{0}[/tex] - C) [tex]e^{-kt}[/tex]
T(5) = 15 + (100 - 15) [tex]e^{-5k}[/tex]
70 = 15 + (85) [tex]e^{-5k}[/tex]
[tex]e^{-5k}[/tex] = (70 - 15) / 85
-5k = ln (55/85)
k = - ln (55/85) / 5
k = 0.087
thus, we have the eqaution
T(t) = 15 + (85) [tex]e^{-0.087t}[/tex]
now we can determine the time when T(t) = 40[tex]^{o}C[/tex]
40 = 15 + (85) [tex]e^{-0.087t}[/tex]
[tex]e^{-0.087t}[/tex] = (40-15)/85
-0.0087t = ln (25/85)
t = - ln (25/85)/0.087
t = 14.07 minutes
