Question

At the moment a hot cake is put in a cooler, the difference between the cake's and the cooler's temperatures is
\[50\degree\] Celsius. This causes the cake to cool and the temperature difference loses

\[\dfrac{1}5\] of its value every minute.
Write a function that gives the temperature difference in degrees Celsius,
\[D(t)\],
\[t\] minutes after the cake was put in the cooler.

Answer 1

Each minute the difference is multiplied by 4/5 (it loses 1/5). With initial difference 50°C,
D(t) = 50(4/5)^t, where t is the time in minutes (t ≥ 0).