Water is being heated in a kettle. At time $t$ seconds, the temperature of the water is $ heta$ °C. The rate of increase of the temperature of the water at any time $t$ is modelled by the differential equation $$\frac{d\theta}{dt} = \lambda(120 - \theta), \quad 0 \leq \theta \leq 100$$ where $ \lambda $ is a positive constant. Given that $ \theta = 20 $ when $ t = 0, $ (a) solve this differential equation to show that $$\theta = 120 - 100 e^{-\lambda t}$$ When the temperature of the water reaches 100 °C, the kettle switches off. (b) Given that $ \lambda = 0.01, $ find the time, to the nearest second, when the kettle switches off.

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Water is being heated in a kettle - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 9

Question 8

Water is being heated in a kettle. At time $t$ seconds, the temperature of the water is $ heta$ °C. The rate of increase of the temperature of the water at any time... show full transcript

Worked Solution & Example Answer:Water is being heated in a kettle - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 9

Step 1

a) Solve this differential equation to show that θ = 120 - 100e^{-λt}

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Answer

To solve the differential equation, we first separate the variables: $\frac{d\theta}{120 - \theta} = \lambda dt$

Integrating both sides, we have: $-\ln(120 - \theta) = \lambda t + C$

Exponentiating yields: $120 - \theta = Ae^{-\lambda t}$

For the integration constant ( A ), we substitute ( \theta = 20 ) when ( t = 0 ):

$120 - 20 = A e^{0} \Rightarrow A = 100$

Now substituting ( A ) back into the equation gives: $\theta = 120 - 100 e^{-\lambda t}$ .

Step 2

b) Given that λ = 0.01, find the time, to the nearest second, when the kettle switches off.

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Answer

When the kettle switches off, the temperature reaches 100 °C:

$\theta = 100$ Substituting into the solved equation: $100 = 120 - 100 e^{-0.01t}$

Rearranging gives: $100 e^{-0.01t} = 20$

Therefore: $e^{-0.01t} = \frac{1}{5}$ Taking natural logs: $-0.01t = \ln(\frac{1}{5})$

Calculating gives: $t = -\frac{\ln(\frac{1}{5})}{0.01} \approx 160.94379 \text{ seconds}$ Rounding to the nearest second: $t \approx 161 \text{ seconds}$ .

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