Water is being heated in an electric kettle - Edexcel - A-Level Maths Pure - Question 5 - 2015 - Paper 3

We know that at t = 40, θ = 70°C. Substitute these values into the equation:

$70 = 120 - 100e^{-40λ}$

Rearranging gives:

$100e^{-40λ} = 120 - 70 = 50$

Dividing by 100:

$e^{-40λ} = 0.5$

Taking the natural logarithm of both sides:

$-40λ = ext{ln}(0.5)$

Thus, λ can be expressed as:

λ = -rac{ ext{ln}(0.5)}{40} = rac{ ext{ln}(2)}{40}

Where a = 2 and b = 40.

We set θ = 100 to find T when the kettle switches off:

$100 = 120 - 100e^{-λT}$

Rearranging gives:

$100e^{-λT} = 120 - 100 = 20$

Now substituting λ:

e^{-λT} = rac{1}{5}

Taking logarithms:

-λT = ext{ln}rac{1}{5}

Substituting for λ:

T = -rac{ ext{ln}rac{1}{5}}{rac{ ext{ln}(2)}{40}} = rac{-40 ext{ln}rac{1}{5}}{ ext{ln}(2)}

Calculating numerically, this results in:

$T ≈ 93$

Thus, the value of T to the nearest whole number is 93.