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Water is being heated in a kettle - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 9

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

Worked Solution & Example Answer:Water is being heated in a kettle - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 9

Step 1

a) solve this differential equation to show that

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Answer

Step 1: Separate Variables

We start with the given differential equation:

dθdt=λ(120θ)\frac{d\theta}{dt} = \lambda(120 - \theta)

We can rearrange this to separate variables:

dθ120θ=λdt\frac{d\theta}{120 - \theta} = \lambda dt

Step 2: Integrate Both Sides

Integrate both sides:

dθ120θ=λdt\int \frac{d\theta}{120 - \theta} = \int \lambda dt

This results in:

ln(120θ)=λt+C-\ln(120 - \theta) = \lambda t + C

Step 3: Solve for θ

By exponentiating both sides, we obtain:

120θ=eλtC120 - \theta = e^{-\lambda t - C}

Letting A=eCA = e^{-C}, we rearrange it to:

θ=120Aeλt\theta = 120 - Ae^{-\lambda t}

Step 4: Apply Initial Condition

Using the initial condition where (\theta = 20) when (t = 0):

20=120AA=10020 = 120 - A \Rightarrow A = 100

Thus, we can rewrite our equation as:

θ=120100eλt\theta = 120 - 100e^{-\lambda t}

This shows the required result.

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

Step 1: Set θ to 100

When the kettle switches off, we set (\theta = 100):

100=120100e0.01t100 = 120 - 100e^{-0.01t}

Step 2: Solve for t

Rearranging gives:

100e0.01t=20100e^{-0.01t} = 20

Now, dividing both sides by 100:

e0.01t=0.2e^{-0.01t} = 0.2

Taking the natural logarithm:

0.01t=ln(0.2)-0.01t = \ln(0.2)

Thus,

t=ln(0.2)0.01t = \frac{-\ln(0.2)}{0.01}

Step 3: Calculate the value

Calculating this we get:

t160.9437t \approx 160.9437

Rounding to the nearest second:

t161t \approx 161

Thus, the kettle switches off at approximately 161 seconds.

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