Question

Water at temperature 100°C cools in 10 minutes to 80°C at a room temperature of 25°C. Find the time when the temperature is 40°C `[log_"e"  11/15 = - 0.3101; log_"e" 5 = 1.6094]`

Answer 1

Let "T" be the temperature in time 't'

Given `"dT"/"dt" oo ("T" - 25)`

⇒ `"d"/"dt" = - "k"("T" - 25)`, k > 0

`int "dT"/("T" - 25) = - "k" int  "dt"`

`log("T" - 25) = - "kt" +"c"`   .........(1)

Given when t = 0, T = 100°C

⇒ c = log 75  ........[∵ From (1)]

(1) ⇒ log(T – 25) = – kt + log 75

`log(("T" - 25)/75)` = – kt

kt = `log(75/("T" - 25))`

Again, given t = 10,  = 80

  ⇒ 10k = `log(75/55)`

10k = `log(15/11)`

∴ k = `1/1 log(5/11)`

= `1/10(0.3101)`

k = 0.03101

∴ 0.03101 t = `log(75/("T" - 25))`

When T = 48, t = ?

⇒ 0.03101 t = `log(75/15)`

0.03101 t = log 5

0.03101 t = 1.6094

t = `1.6094/0.03101`

= 52 minutes