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<p>A freshly brewed cup of tea, initially at a temperature of ${85^\circ\text{C}}$, is left out to cool in a room with a room temperature of ${25^\circ\text{C}}$. After ${20}$ minutes, the temperature of the tea is ${55^\circ\text{C}}$.</p>
<p>It is known that the rate of change of the temperature of the tea is proportional to the difference between the temperature of the tea and the room temperature.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Given that&nbsp;$\theta$ is the temperature of the tea $t$ minutes after the tea is left out to cool, show that&nbsp;$\theta = 25 + 60\mathrm{e}^{-kt}$, where&nbsp;$k$ is a constant to be determined.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Sketch the graph of $\theta$ against $t$.<span class="inline-marks">[2]</span></div></li></ol>

A freshly brewed cup of tea, initially at a temperature of ${85^\circ\text{C}}$, is left out to cool in a room with a room temperature of ${25^\circ\text{C}}$. After ${20}$ minutes, the temperature of the tea is ${55^\circ\text{C}}$.

It is known that the rate of change of the temperature of the tea is proportional to the difference between the temperature of the tea and the room temperature.

Solving Differential Equations by Separation of Variables

Newton’s Law of Cooling and Warming

Sketching Solution Curves of Differential Equations

2025 ACJC P1 Q7

Differential Equations

<p>A freshly brewed cup of tea, initially at a temperature of $85$$^\circ C$, is left out to cool in a room with a room temperature of $25$$^\circ C$. After $20$ minutes, the temperature of the tea is $55$$^\circ C$. It is known that the rate of change of the temperature of the tea is proportional to the difference between the temperature of the tea and the room temperature.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Given that $\theta$ is the temperature of the tea $t$ minutes after the tea is left out to cool, show that $\theta = 25 + 60\mathrm{e}^{-kt}$, where $k$ is a constant to be determined.</span><span class="inline-marks">[5]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Sketch the graph of $\theta$ against $t$.</span><span class="inline-marks">[2]</span></div></li></ol>

A freshly brewed cup of tea, initially at a temperature of $85$$^\circ C$, is left out to cool in a room with a room temperature of $25$$^\circ C$. After $20$ minutes, the temperature of the tea is $55$$^\circ C$. It is known that the rate of change of the temperature of the tea is proportional to the difference between the temperature of the tea and the room temperature.

Given that $\theta$ is the temperature of the tea $t$ minutes after the tea is left out to cool, show that $\theta = 25 + 60\mathrm{e}^{-kt}$, where $k$ is a constant to be determined.

Sketch the graph of $\theta$ against $t$.

Techniques of Differentiation

<p>Alice bakes a large pie and removes it from the oven at $1$pm. The temperature reading of the pie using a food thermometer is $200{}^\circ \mathrm{C}$ at that instant.</p>
<p>She left the pie on the kitchen countertop to cool down. The temperature of the kitchen is kept at a constant $32{}^\circ \mathrm{C}$. The temperature of the pie $t$ minutes after it is removed from the oven is $\theta {}^\circ \mathrm{C}$. This temperature decreases at a rate proportional to the difference between the temperature of the pie and the temperature of the kitchen.</p>
<p>At $1.15$pm, Alice checks the temperature of the pie again and it is $180{}^\circ \mathrm{C}$.</p>
<ol class="abc-list"><li data-list-item-id="item-1760697028857-cvyzfl505" data-list-item-path="item-1760697028857-cvyzfl505"><div class="list-item-paragraph">Find an exact expression for&nbsp;$\theta $ in terms of $t$, simplifying your answer.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="item-1760697029352-skmh73k68" data-list-item-path="item-1760697029352-skmh73k68"><div class="list-item-paragraph">Sketch a graph of $\theta $ against $t$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760697030747-btxgjonc1" data-list-item-path="item-1760697030747-btxgjonc1"><div class="list-item-paragraph">Food safety experts caution that bacteria grow fastest in food within the temperature range of $5{}^\circ \mathrm{C}$ to $60{}^\circ \mathrm{C}$, known as the temperature danger zone. Find the latest time to the nearest minute to safely store the pie to avoid this danger zone.<span class="inline-marks">[2]</span></div></li></ol>

Alice bakes a large pie and removes it from the oven at $1$pm. The temperature reading of the pie using a food thermometer is $200{}^\circ \mathrm{C}$ at that instant.

She left the pie on the kitchen countertop to cool down. The temperature of the kitchen is kept at a constant $32{}^\circ \mathrm{C}$. The temperature of the pie $t$ minutes after it is removed from the oven is $\theta {}^\circ \mathrm{C}$. This temperature decreases at a rate proportional to the difference between the temperature of the pie and the temperature of the kitchen.

At $1.15$pm, Alice checks the temperature of the pie again and it is $180{}^\circ \mathrm{C}$.

2023 RI P1 Q6

<p>Scientists are investigating how the temperature of water changes in various environments.</p>
<ol class="i-list"><li><div class="list-item-paragraph">The scientists begin by investigating how hot water cools.</div><div class="list-item-paragraph">The water is heated in a container and then placed in a room which is kept at a constant temperature of $16\,{}^\circ \mathrm{C}$. The temperature of the water $t$ minutes after it is placed in the room is $\theta \,{}^\circ \mathrm{C}\mathrm{.}$ This temperature decreases at a rate proportional to the difference between the temperature of the water and the temperature of the room. The temperature of the water falls from a value of $80\,{}^\circ \mathrm{C}$ to $32\,{}^\circ \mathrm{C}$ in the first $30$ minutes.</div><ol class="abc-list"><li><div class="list-item-paragraph">Write down the differential equation for this situation. Solve this differential equation to get $\theta $ as an exact function of $t$.</div><p style="text-align: right;">[6]</p></li><li><div class="list-item-paragraph">Find the temperature of the water $45$ minutes after it is placed in the room.</div><p style="text-align: right;">[1]</p></li></ol></li><li><div class="list-item-paragraph">The scientists then model the thickness of ice on a pond.</div><div class="list-item-paragraph">In winter the surface of the water in the pond freezes. Once the thickness of the ice reaches $3$ cm, it is safe to skate on the ice. The freezing of the water is modelled by a differential equation in which the rate of change of the thickness of ice is inversely proportional to its thickness. It is given that $T=0$ when $t=0.$ After $60$ minutes, the ice is $1$ cm thick.</div><div class="list-item-paragraph">Find the time from when freezing commences until the ice is first safe to skate on.</div><p style="text-align: right;">[6]</p></li></ol>

Scientists are investigating how the temperature of water changes in various environments.

Modelling with Differential Equations

N2019 P1 Q11

<ol class="abc-list" start="2"><li><div class="list-item-paragraph">A freshly brewed cup of coffee has an initial temperature $95{}^\circ C$. It is placed in a room where the temperature is a constant at $20{}^\circ C$. As the coffee cools down, the rate of decrease of its temperature $\theta {}^\circ C$ after time $t$ minutes is proportional to the temperature difference $\left( \theta -20 \right){}^\circ C$. It is cooling at a rate of $1{}^\circ C$ per minute when its temperature is $70{}^\circ C$.</div><ol class="i-list"><li><div class="list-item-paragraph">By setting up a differential equation, find the rate of cooling of the coffee when its temperature is $40{}^\circ C$ and show that $\theta =20+75{{e}^{-0.02t}}$ .</div></li><li><div class="list-item-paragraph">Comment on whether the model can be regarded as a good model of the situation in the real world.</div></li></ol></li></ol>

Assumptions and Limitations of Differential Equation Models

2014 NJC P1 Q11

<p>When a cake is removed from the oven, its temperature decreases at a rate proportional to the positive difference between its temperature and the temperature of the room. The temperature of the room is constant at $25{}^\circ C$ and $T$ is the temperature of the cake $t$ hours after removing from the oven. The temperature of the cake at the instant when it is removed from the oven is $50{}^\circ C$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Show that $T=25+25{{e}^{-kt}}$ , where $k$ is a positive constant.<span class="inline-marks">[5]</span></div></li><li><div class="list-item-paragraph">Sketch a solution curve for $T$ where $t\ge 0$ .<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Pradip is a world-renowned baker who is uncompromising on the freshness of the cakes served by his bakery. On a particular morning at 10.00am, he found the temperature of a cake at $40{}^\circ C$, and $1$ hour later, it was $31{}^\circ C$. At what time, to the nearest minute, was the cake removed from the oven?<span class="inline-marks">[4]</span></div><div class="list-item-paragraph">State what happens to the temperature of the cake in the long run.<span class="inline-marks">[1]</span></div></li></ol>

When a cake is removed from the oven, its temperature decreases at a rate proportional to the positive difference between its temperature and the temperature of the room. The temperature of the room is constant at $25{}^\circ C$ and $T$ is the temperature of the cake $t$ hours after removing from the oven. The temperature of the cake at the instant when it is removed from the oven is $50{}^\circ C$.

Equilibrium Solutions and Long-Term Behaviour

2013 MI P1 Q9

<p>The rate of temperature loss of an animal corpse can be estimated using Newton’s Law of Cooling, which states that the rate of change of temperature $\theta {}^\circ \mathrm{C}$, $t$ hours after death of an animal is proportional to the difference between its body temperature $\theta {}^\circ \mathrm{C}$ and the constant surrounding temperature ${{\theta }_{0}}{}^\circ \mathrm{C}$, where&nbsp;$\theta &gt;{{\theta }_{0}}$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Write down a differential equation for this situation. Solve this differential equation and show that the general solution of the above differential equation is given by $\theta ={{\theta }_{0}}+A{{e}^{-kt}}$, where&nbsp;$A$ and $k$ are positive constants.<span class="inline-marks">[3]</span></div></li></ol>
<p>It is given that&nbsp;${{\theta }_{0}}=24$, the initial value of $\theta $ is $36$ and the initial rate of temperature loss is $2.5{}^\circ \mathrm{C}$ per hour.</p>
<ol class="i-list" start="2"><li><div class="list-item-paragraph">Calculate the exact values of $A$ and $k$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Hence, sketch the graph of $\theta $ against the $t$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Explain why the rate of change of temperature of an animal corpse cannot be modelled by a constant rate of decrease of $1.5{}^\circ \mathrm{C}$ per hour.<span class="inline-marks">[1]</span></div></li></ol>

The rate of temperature loss of an animal corpse can be estimated using Newton’s Law of Cooling, which states that the rate of change of temperature $\theta {}^\circ \mathrm{C}$, $t$ hours after death of an animal is proportional to the difference between its body temperature $\theta {}^\circ \mathrm{C}$ and the constant surrounding temperature ${{\theta }_{0}}{}^\circ \mathrm{C}$, where $\theta >{{\theta }_{0}}$.

It is given that ${{\theta }_{0}}=24$, the initial value of $\theta $ is $36$ and the initial rate of temperature loss is $2.5{}^\circ \mathrm{C}$ per hour.

2022 SAJC P1 Q7

<p>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 $\left( 20-\theta \right){}^\circ \mathrm{C}$. 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;$\theta =20-10{{e}^{-\frac{1}{10}t}}$<span class="inline-marks">[7]</span></p>
<p>Find the time it takes 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 $t$.<span class="inline-marks">[4]</span></p>

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 $\left( 20-\theta \right){}^\circ \mathrm{C}$. 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 $\theta =20-10{{e}^{-\frac{1}{10}t}}$

Find the time it takes 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 $t$.

N2010 P1 Q7

<p>A tank contains water which is heated by an electric water heater working under the action of a thermostat. When the water heater is first switched on, the temperature of the water is $35\mathrm{ }{}^\circ \mathrm{C}$. The heater causes the temperature to increase at a rate $r\mathrm{ }{}^\circ \mathrm{C}$ per minute, where $r$ is a constant, until the water temperature hits $75\mathrm{ }{}^\circ \mathrm{C}$. The heater then switches off.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Write down, in terms of $r$, the time taken for the temperature to increase from $\mathrm{35 }{}^\circ \mathrm{C}$ to $75\mathrm{ }{}^\circ \mathrm{C}$.</span></div><p style="text-align: right;">[1]</p></li></ol>
<p>The temperature of the water then immediately starts to decrease. The temperature of the water at time $t$ minutes after the heater is switched off is $\theta \mathrm{ }{}^\circ \mathrm{C}$. It is known that the temperature of the water decreases at a variable rate $k\left( \theta -25 \right)\mathrm{ }{}^\circ \mathrm{C}$ per minute, where $k$ is a positive constant, until $\theta =35$.</p>
<ol class="i-list" start="2"><li><div class="list-item-content"><span class="list-item-text">Write down a differential equation involving $\theta $ and $t$, to represent the situation as the temperature is decreasing.</span></div><p style="text-align: right;">[1]</p></li><li><div class="list-item-content"><span class="list-item-text">Given that when $\theta =55$, the temperature is decreasing at a rate of $5\mathrm{ }{}^\circ \mathrm{C}$ per minute, find the total length of time for the temperature to increase from $35\ {}^\circ \mathrm{C}$ to $75\ {}^\circ \mathrm{C}$ and then decrease to $35\ {}^\circ \mathrm{C}$, leaving your answer in exact form, in terms of $r$.</span></div><p style="text-align: right;">[7]</p></li></ol>

A tank contains water which is heated by an electric water heater working under the action of a thermostat. When the water heater is first switched on, the temperature of the water is $35\mathrm{ }{}^\circ \mathrm{C}$. The heater causes the temperature to increase at a rate $r\mathrm{ }{}^\circ \mathrm{C}$ per minute, where $r$ is a constant, until the water temperature hits $75\mathrm{ }{}^\circ \mathrm{C}$. The heater then switches off.

Write down, in terms of $r$, the time taken for the temperature to increase from $\mathrm{35 }{}^\circ \mathrm{C}$ to $75\mathrm{ }{}^\circ \mathrm{C}$.

The temperature of the water then immediately starts to decrease. The temperature of the water at time $t$ minutes after the heater is switched off is $\theta \mathrm{ }{}^\circ \mathrm{C}$. It is known that the temperature of the water decreases at a variable rate $k\left( \theta -25 \right)\mathrm{ }{}^\circ \mathrm{C}$ per minute, where $k$ is a positive constant, until $\theta =35$.

Write down a differential equation involving $\theta $ and $t$, to represent the situation as the temperature is decreasing.

Given that when $\theta =55$, the temperature is decreasing at a rate of $5\mathrm{ }{}^\circ \mathrm{C}$ per minute, find the total length of time for the temperature to increase from $35\ {}^\circ \mathrm{C}$ to $75\ {}^\circ \mathrm{C}$ and then decrease to $35\ {}^\circ \mathrm{C}$, leaving your answer in exact form, in terms of $r$.

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