Joan brings a cup of hot tea into a room and places the cup on a table - Edexcel - A-Level Maths Pure - Question 5 - 2011 - Paper 4

We need to find the value of k. Given that it takes 5 minutes to decrease from 90°C to 55°C:

At t = 5, θ = 55:

$55 = 20 + 70e^{-5k}$

This leads to:

$55 - 20 = 70 e^{-5k}$

We simplify this to:

$35 = 70 e^{-5k}$

Dividing both sides by 70 gives:

$\frac{1}{2} = e^{-5k}$

Taking natural logs:

$ext{ln}(\frac{1}{2}) = -5k$

And since ln(\frac{1}{2}) = -\ln 2, we have:

$-\ln 2 = -5k$

Thus,

$k = \frac{1}{5} \ln 2$

To find the rate of change of temperature with respect to time, we differentiate the temperature equation:

$θ = 20 + 70e^{-kt}$

Differentiating with respect to t:

$\frac{dθ}{dt} = -70ke^{-kt}$

Substituting k we found earlier:

$k = \frac{1}{5} \ln 2$

At t = 10:

$\frac{dθ}{dt} = -70 \times \frac{1}{5} \ln 2 \times e^{-\frac{1}{5} \ln 2 \cdot 10}$

This simplifies to:

$\frac{dθ}{dt} = -14 \ln 2 e^{-2 \ln 2} = -14 \cdot \frac{1}{4} = -3.5$

Thus the rate of decrease of θ is approximately:

$-3.5$

=> Rate of decrease of temperature = 2.426 °C/min (to 3 decimal places).