9 The temperature of a freezer is \(- 20 ^ { \circ } \mathrm { C }\). A container of a liquid is placed in the freezer. The rate at which the temperature, \(\theta ^ { \circ } \mathrm { C }\), of a liquid decreases is proportional to the difference in temperature between the liquid and its surroundings. The situation is modelled by the differential equation $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k ( \theta + 20 ) ,$$ where time \(t\) is in minutes and \(k\) is a positive constant.

1. Express \(\theta\) in terms of \(t , k\) and an arbitrary constant. Initially the temperature of the liquid in the container is \(40 ^ { \circ } \mathrm { C }\) and, at this instant, the liquid is cooling at a rate of \(3 ^ { \circ } \mathrm { C }\) per minute. The liquid freezes at \(0 ^ { \circ } \mathrm { C }\).
2. Find the value of \(k\) and find also the time it takes (to the nearest minute) for the liquid to freeze. The procedure is repeated on another occasion with a different liquid. The initial temperature of this liquid is \(90 ^ { \circ } \mathrm { C }\). After 19 minutes its temperature is \(0 ^ { \circ } \mathrm { C }\).
3. Without any further calculation, explain what you can deduce about the value of \(k\) in this case.