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