A student is conducting an experiment in a laboratory to investigate how quickly liquids cool to room temperature - AQA - A-Level Maths Pure - Question 8 - 2019 - Paper 3

First, substitute θ = 68 °C and t = 2 into the equation:

$68 = 5(4 + λe^{-2k})$

Rearranging gives:

$68 = 20 + 5λe^{-2k}$

This simplifies to:

$48 = 5λe^{-2k}$

Thus:

λe^{-2k} = rac{48}{5} = 9.6

Next, we need to express λ in terms of k. We do this by substituting our initial temperature at t = 0:

$75 = 5(4 + λe^{0})$

This leads to:

$75 = 20 + 5λ$

So:

ightarrow λ = 11$$ Now substituting into the earlier equation gives: $$11e^{-2k} = 9.6$$ From which: $$e^{-2k} = rac{9.6}{11} ightarrow e^{-2k} = 0.872727272727$$ Taking the natural logarithm: $$-2k = ext{ln}(0.872727272727)$$ And calculating k: $$k = -rac{1}{2} ext{ln}(0.872727272727) ≈ 0.068066$$