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. What was the temperature of the coffee at 10.15 AM? `|log  9/100 = - 0.6061|`

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)

When t = 15, T = ?

(2) ⇒ T = 110 e^k(15) + 70

T = `110(9/11)^(15/10) + 70`

T = `110(9/11)^(3/2) + 70`

T = `110 xx (9/110) sqrt(9/11) + 70`

T = `110 xx 9/11 xx 3/sqrt(11) + 70`

T = `270/sqrt(11) + 70`

T= `270/3.32 + 70`

T = 81.33 + 70

T = 151.3°F

∴ The temperature of the coffee at 10.15 A.M is 151.3°F