9 Newton's law of cooling states that the rate at which the temperature of an object is falling at any instant is proportional to the difference between the temperature of the object and the temperature of its surroundings at that instant. A container of hot liquid is placed in a room which has a constant temperature of \(20 ^ { \circ } \mathrm { C }\). At time \(t\) minutes later, the temperature of the liquid is \(\theta ^ { \circ } \mathrm { C }\).

1. Explain how the information above leads to the differential equation $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k ( \theta - 20 ) ,$$ where \(k\) is a positive constant.
2. The liquid is initially at a temperature of \(100 ^ { \circ } \mathrm { C }\). It takes 5 minutes for the liquid to cool from \(100 ^ { \circ } \mathrm { C }\) to \(68 ^ { \circ } \mathrm { C }\). Show that $$\theta = 20 + 80 \mathrm { e } ^ { - \left( \frac { 1 } { 5 } \ln \frac { 5 } { 3 } \right) t }$$
3. Calculate how much longer it takes for the liquid to cool by a further \(32 ^ { \circ } \mathrm { C }\).