6. Water is being heated in a kettle. At time \(t\) seconds, the temperature of the water is \(\theta ^ { \circ } \mathrm { C }\). The rate of increase of the temperature of the water at any time \(t\) is modelled by the differential equation $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = \lambda ( 120 - \theta ) , \quad \theta \leqslant 100$$ where \(\lambda\) is a positive constant. Given that \(\theta = 20\) when \(t = 0\),

1. solve this differential equation to show that $$\theta = 120 - 100 \mathrm { e } ^ { - \lambda t }$$ When the temperature of the water reaches \(100 ^ { \circ } \mathrm { C }\), the kettle switches off.
2. Given that \(\lambda = 0.01\), find the time, to the nearest second, when the kettle switches off.