smc-1184-50-Cooling problems

A container of water is heated to boiling point (100°C) and then placed in a room that has a constant temperature of 20°C. After five minutes the temperature of the water is 80°C.

Use Newton’s law of cooling `(dT)/(dt) = -k (T - 20)`, where `T text(°C)` is the temperature of the water at the time `t` minutes after the water is placed in the room, to show that `e^(-5k) = 3/4.` (2 marks)
Find the temperature of the water 10 minutes after it is placed in the room. (3 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`65`

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a.	`(dT)/(dt)`	`= -k (T – 20)`
	`(dt)/(dT)`	`= -1/k * 1/(T – 20)`
	`t`	`= -1/k int 1/(T – 20)\ dT`
	`-kt`	`= log_e (T – 20) + c`

`text(When)\ \ t = 0,\ \ T = 100,`

`=>c = -log_e 80,`

`:. -kt`	`= log_e (T – 20) – log_e 80`
	`= log_e ((T – 20)/80)`

`text(When)\ \ t = 5,\ \ T = 80,`

`-5k = log_e (3/4)`

`:. e^(-5k) = 3/4`

(b) `text(Find)\ \ T\ \ text(when)\ \ t = 10:`

`-10k`	`= log_e ((T – 20)/80)`
`(T – 20)/80`	`= e^(-10k)`
`(T – 20)/80`	`= (e^(-5k))^2`
`(T – 20)/80`	`= (3/4)^2`
`:. T`	`= (3/4)^2 xx 80 + 20`
	`=(9 xx 80)/16 +20`
	`= 65^@ text(C)`