Mathematical modelling

Question

A cup of coffee at C

o

90 is kept in a room at C

o

25 . Two minutes later, the

temperature of the coffee is C

o

75 . When will the temperature of the coffee reach

C

o

50 ?

Answer 1

How about this:

We know that cooling occurs via an exponential function. So the basic function will be:

f(t) = A(e^rt) In this case r (the time constant) will be negative because it's cooling, so f(t) = A(e^-rt)

We know the initial temp is 90 and the steady state (t = big numbers) temperature is 25 (it will cool to room temp). So, because the horizontal asymptote to f(t) = A(e^-rt) is zero, we can adjust it upwards to 25 by making the equation be f(t) = A(e^-rt + 25/A). That way when t gets big and e^-rt approaches zero, f(t) = A(0 + 25/A) or 25.

So, using f(t) = A(e^-rt + 25/A), we know f(0) = 90 so 90 = A(e^-r(0) + 25/A) = A + 25 or A = 65.

Now using f(t) = 65(e^-rt + 25/65) and f(2) = 75, we can calculate r:

75 = 65(e^-r(2) + 25/65)

75/65 = e^-2r + 25/65

50/65 = e^-2r

ln(50/65) = -2r

r = 0.1311821322

So the modeling function becomes: f(t) = 65(e^{-0.1311821322t} + 25/65)

To find t for a temperature of 50, plug in 50 for f(t) and solve:

50 = 65(e^{-0.1311821322t} + 25/65)

0.76923 = e^{-0.1311821322t} + 25/65

0.38462 = e^{-0.1311821322t}

ln(0.38462) = -0.1311821322t

t = ln(0.38462)/(-0.1311821322) = 7.28 minutes