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Tamil Nadu Board of Secondary EducationHSC Science Class 12

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 - Mathematics

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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

Sum
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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

shaalaa.com
Applications of First Order Ordinary Differential Equations
  Is there an error in this question or solution?
Chapter 10: Ordinary Differential Equations - Exercise 10.8 [Page 174]

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Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 12 TN Board
Chapter 10 Ordinary Differential Equations
Exercise 10.8 | Q 9 | Page 174

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