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Question
A pot of boiling water at 100°C is removed from a stove at time t = 0 and left to cool in the kitchen. After 5 minutes, the water temperature has decreased to 80° C and another 5 minutes later it has dropped to 65°C. Determine the temperature of the kitchen
Solution
Let T be the temperature of the boiling water.
Tm is the temperature of the kitchen.
By Newton’s law of cooling, we get
`"dT"/"dt" = "k"("T" - "T"_"m")`
The equation can be written as
Taking Integration on both sides, we get
`int "dt"/("T" - "T"_"M") = "k" int "dt"`
⇒ `log("T" - "T"_"m") = "kt" + log"c"`
⇒ `log("T" - "T"_"m") - log "c" = "kt"`
`log(("T" - "T"_"m")/"c")` = kt
`("T" - "T"_"m")/"c" "e"^("kt")`
`"T" - "T"_"m" = "ce"^"kt"` ........(1)
Initial condition:
When t = 0, T = 100
`100 - "T"_"m" = "ce"^("k"(0))`
`100 - 'T"_"m" = "ce"^0`
`100 - "T"_"m"` = c
c = `100 - "T"_"m"`
Substituting the value of c in equation (1), we get
`"T" - "T"_"m" = (1 - "T"_"m")"e"^"kt"` ........(2)
Also given when t = 5, T = 80
`80 - "T"_"m" = (100 - "T"_"m") "e"^(5"k")`
`(80 - "T"_"m")/(100 - "T"_"m") = "e"^(5"k")`
∴ `"e"^(5"k") = (80 - "T"_"m")/(100 - "T"_"m")` .......(3)
After that when t = 10, T = 65
(2) ⇒ `65 - "T"_"m"(100 - "T"_"m") "e"^(10"k")`
`(65 - "T"_"m")/(100 - "T"_"m") = "e"^(10"k")`
`"e"^(10"k") = (65 - "T"_"m")/(100 - "T"_"m")`
`("e"^(5"k"))^2 = (65 - "T"_"m")/(100 - "T"_"m")`
`((80 - "T"_"m")/(100 - "T"_"m"))^2 = (65 - "T"_"m")/(100 - "T"_"m")`
`(80 - "T"_"m")^2/(100 - "T"_"m")^2 = (65 - "T"_"m")/(100 - "T"_"m")`
`(80 - "T"_"m")^2/(100 - "T"_"m") = 65 - "T"_"m"`
`(80 - "T"_"m")^2 = (65 - "T"_"m")(100 - "T"_"m")`
(a – b)2 = a2 + b2 – 2ab
a = 80, b = Tm
`6400 + "T"_"m"^2 - 160 "T"_"m" = 6500 - 100"T"_"m" - 65"T"_"m" + "T"_"m"^2`
6400 – 6500 = 160 Tm – 165 Tm
– 100 = – 5Tm
Tm = `100/5`
Tm = 20°C
Hence the temperature of the kitchen be 20°C
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