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Calculus, SPEC1 2013 VCAA 5

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.

  1. 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)

  2. Find the temperature of the water 10 minutes after it is placed in the room.   (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `65`
Show Worked Solution
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)`