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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]`
Solution
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
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