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

the engine cool to 40
`^(o)C` at 14.07 minutes

Step-by-step 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