Question

The temperature of a cup of coffee cools from 105 deg to room temperature (20 deg). After 5 mins the temperature is 95 deg. Find (a) formula for the temperature at time t, (b) the temperature after 11 mins c) time for the temperature to drop to 85 deg?

Answer 1

(a) `T = 20 + 85(15/17)^(1/5t)`

(b) `~~ 85` °C

(c) After `17.4` minutes

Explanation:

I have a bit of an issue with the way this question is worded, as you must be expected to know the relationship that models the situation. Without this inner knowledge one could assume it was a linear relationship between the temperature and time which doesn't sound unreasonable.

In fact the relationship is defined by Newtons Law of Cooling which states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature (i.e. the temperature of its surroundings).

Let us define the following variables:

`{ (T,"Temperatue of the cup of coffee (°C)"), (t, "time (minutes)") :}`

Then, as the cup of coffee is cooling down, using Newtons Law of Cooling we have:

`-(dT)/dt prop T-20 => (dT)/dt = -k(T-20)`

where `k` is the constant of proportionality. This is a First Order separable Differential Equation and we can separate the variables to get:

`int \ 1/(T-20) \ dT = int \ -k \ dt`

Which we can integrate to get:

`\ \ \ ln |T-20| = -kt + C`

NB: we can drop the modulus as the coffee temperature will not drop below room temperature, and hence `T-20 > 0`

We initially started off with `T=105` (°C) `=>T=105` when `t=0`, so we can substitute into the DE solution to get:

`ln (105-20) = 0 + C => C = ln85`

We are also told that `T=95` (°C) when `t=5` so we can substitute into the DE solution to get:

`ln (95-20) = -5k + ln85 => 5k = ln85-ln75`
`:. k = 1/5ln(85/75)`
`:. k = 1/5ln(17/15)`

(a) And so the Specific Solution is:

`\ \ \ \ \ ln (T-20) = -1/5ln(17/15)t + ln85` ..... [1]
`:. ln (T-20) = ln{(17/15)^(-1/5t)} + ln85`
`:. ln (T-20) = ln{(15/17)^(1/5t)} + ln85`
`:. ln (T-20) = ln{85(15/17)^(1/5t)}`
`:. " "T-20 = 85(15/17)^(1/5t)`
`:. " " T = 20 + 85(15/17)^(1/5t)`

[ We should just check that we have not made a mistake by checking the initial conditions:

`t=0 => T=20 + 85(15/17)^0 = 105`
`t=5 => T=20+85(15/17)^1 = 95`

so we know the solution is sound]

(b) We are asked to find `T` when `t=11`, and so we have:

`T = 20 + 85(15/17)^(11/5)`
`\ \ \ = 84.540461 ....`
`\ \ \ ~~ 85` °C

(c) We are asked to find `t` when `T=75`, and so using [1] we have:

`" "ln (75-20) = -1/5ln(17/15)t + ln85`
`:. " "ln (55) = -1/5ln(17/15)t + ln85`
`:. 1/5ln(17/15)t = ln85 -ln55`
`:. 1/5ln(17/15)t = ln(85/55)`
`:. " "t = 5ln(17/11)/ln(17/15)`
`:. " "t = 17.390026...`
`:. " "t = 17.4` minutes