Question

At 10.00 A.M. a woman took a cup of hot instant coffee from her microwave oven and placed it on a nearby kitchen counter to cool. At this instant, the temperature of the coffee was 180°F and 10 minutes later it was 160°F. Assume that the constant temperature of the kitchen was 70°F. The woman likes to drink coffee when its temperature is between130°F and 140°F. between what times should she have drunk the coffee? `|log  6/11 =  - 0.2006|`

Answer 1

Let T be the temperature of a coffee at time t.

T_k be the temperature of the kitchen

By Newton’s law of cooling

`"d"/"dt" = "k"("T" - "T"_"k")`

Given T_k =70

`"dT"/"dt" = "k"("T" - 70)`

The equation can be written as 

`"dT"/("T" - 70)` k dt

Taking integration on both sides,

`int "dT"/("T" - 70) = int "k"  "dt"`

`log("T" - 70) = int "dt"`

`lg("T" - 70)` = kt + log c

`log("T" - 70) - log "c"` = kt

`log(("T" - 70)/"c")` = kt

`("T" - 70)/"c" = "e"^"kt"`

`"T" - 70 = "ce"^"kt"`

T = `"ce"^"kt" + 70`   ........(1) 

Initial condition:

When t = 0, T = 180°F

180 = ce^k(0) + 70

180 = ce° + 70

180 – 70 = c

∴ c = 110°

Substituting c value in equation (1), we get

T = ce^+kt + 70

T = 100 e^kt + 70 ........(2)

Second condition:

when t = 10, T = 160

(2) ⇒ 160 = 110 e^10k + 70

160 – 70 = 110 e^10k 

`90/10` = e^10k 

`9/11` = e^10k 

e^k = `(9/11)^(1/10)`   .........(3)

Woman’s like to drink a coffee between 130°F and 140°F.

(a) when T = 130°F

(2) ⇒ T = 110 e^kt + 70

130 – 70 = 110 e^kt 

`60/110 = 6/11` = e^kt 

e^{kt =} `6/11`

`(9/11)^("t"/10) = 6/11`  ........(By (3))

Taking g on both sides, we get

`log(9/11)^("t"/10) = log  6/11`

`"t"/10 log(9/11) = log  6/11`

`"t"/10 = (log(6/11))/(log(9/11))`

`"t"/10 = (log(0.55))/(log(0.818))`

= `(- 0264)/(- 0.087)`

= 3.0345

t = 10 × 3.0345

= 30.345

(b) When T = 140°

(2) ⇒ T = 110 e^kt + 70

160 – 70 = 110 e^kt 

`70/110` = e^kt 

`7/11` = e^kt 

e^kt = `7/11`

`(9/11)^("t"/10) =  7/11`  ........(By (3))

Taking log on both sides, we get

`log(9/10)^("t"/10) = log  7/11`

`"t"/10 (9/11) = log  7/11`

`"t"/10 = (log(7/11))/(log  9/11)`

= `(- 0.197)/(- 0.087)`

= 2.264

t = `10 xx 2.264`

t = 22.6 min

She drinks coffee between 10.22 and 10.30 approximately.