Question

When a cold drink is taken from a refrigerator, its temperature is 5 degrees C. After 25 minutes in a 20 degrees C room its temperature has increased to 10 degrees C.

(a) What is the temperature of the drink after 50 minutes?
(b) When will its temperature by 15 degrees C?

Answer 1

Using Newton's Law of Cooling,
`(dT)/(dt) = k(T-T_s)`, we have
`(dT)/(dt) = k(T - 20)` (
`T_s` is 20 degrees). Letting
`y = T - 20`, we get
`(dy)/(dt) = ky`, so
`y(t) = y(0)e^(kt)`.
`y(0) = T(0) - 20 = 5-20 = -15`, so
`y(25) = y(0) e^(25k) = -15e^(25k)`, and
`y(25) = T(25) - 20 = 10 - 20 = -10` so
`-15e^(25k) = -10 \ \ \Rightarrow \ \ e^(25k) = (2)/(3)`. Thus,
`25k = \ln\left((2)/(3)\right)` and
`k = (1)/(25) \ln\left((2)/(3)\right)`, so
`y(t) = y(0)e^(kt) = -15e^((1/25)\ln(2/3)t)`. More simply,

`e^(25k) = (2)/(3) \ \ \Rightarrow \ \ e^k = \left( (2)/(3) \right)^(1/25)\\ \Rightarrow \ \ e^(kt) = \left( (2)/(3) \right)^(t/25)\\ \Rightarrow y(t) = -15 \cdot \left( (2)/(3)\right)^(t/25)`

(a)
`T(50) = 20 + y(50) = 20-15 \cdot \left( (2)/(3)\right)^(50/25)` =

`20 - 15 \cdot\left((2)/(3)\right)^2 = 20 - (20)/(3) = 13.\overline{3}{}^(\circ)C`

13.33333 °C

(b)
`15 = T(t) = 20 + y(t) = 20 - 15\cdot\left( (2)/(3) \right)^(t/15)`

`\Rightarrow 15 \cdot \left((2)/(3)\right)^(t/25) = 5 \Rightarrow \left( (2)/(3) \right)^(t/25) = (1)/(3) \Rightarrow`

`(t/25) \ln\left( (2)/(3)\right) = \ln\left( (1)/(3)\right) \rightarrow t = 25 \ln\left( (1)/(3)\right) / \ln\left( (2)/(3)\right) \approx 67.74\text{ min}`

67.74 min