A bottle containing liquid is taken from a refrigerator and placed in a room where the temperature is a constant \(20^\circ\mathrm{C}\). As the liquid warms up, the rate of increase of its temperature \(\theta^\circ\mathrm{C}\) after time \(t\) minutes is proportional to the temperature difference \((20 - \theta)^\circ\mathrm{C}\). Initially the temperature of the liquid is \(10^\circ\mathrm{C}\) and the rate of increase of the temperature is \(1^\circ\mathrm{C}\) per minute. By setting up and solving a differential equation, show that&nbsp;\(\theta = 20 - 10e^{-\frac{1}{10}t}\).[7]
Find the time it take the liquid to reach a temperature of \(15^\circ\mathrm{C}\), and state what happens to \(\theta\) for large values of \(t\). Sketch a graph of \(\theta\) against \(t\).[4]

A bottle containing liquid is taken from a refrigerator and placed in a room where the temperature is a constant \(20^\circ\mathrm{C}\). As the liquid warms up, the rate of increase of its temperature \(\theta^\circ\mathrm{C}\) after time \(t\) minutes is proportional to the temperature difference \((20 - \theta)^\circ\mathrm{C}\). Initially the temperature of the liquid is \(10^\circ\mathrm{C}\) and the rate of increase of the temperature is \(1^\circ\mathrm{C}\) per minute. By setting up and solving a differential equation, show that \(\theta = 20 - 10e^{-\frac{1}{10}t}\).

Find the time it take the liquid to reach a temperature of \(15^\circ\mathrm{C}\), and state what happens to \(\theta\) for large values of \(t\). Sketch a graph of \(\theta\) against \(t\).

2018 TGM P1 Q11

2018 TGM P1 Q11

Solution:

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