Science:Math Exam Resources/Courses/MATH100/December 2019/Question 09(a)

MATH100 December 2019

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Question 09(a)
You put a bottle of room temperature water into your refrigerator. The temperature of your apartment is $23^{\circ }$C, while the temperature of your refrigerator is $3^{\circ }$C. After $10$ minutes the water has cooled to $13^{\circ }$C. Give a formula for the temperature of the water $t$ minutes after it has been placed in the refrigerator.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
This sort of problem is modelled by Newton's law of cooling. Let $T(t)$ be the temperature of the water at time $t$. Can you write down a relationship between $T'(t)$, $T(t)$ and possibly some constants?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let $T(t)$ be the temperature of the water $t$ minutes after putting it in the fridge. By Newton's law of cooling, the following equation holds $T'(t)=k(T(t)-E),$ where $k$ is some constant and $E$ is the temperature of the surroundings. In our case, $E=3$. Note that in the situation we're modelling, $T(t)$ is larger than $E$ and the temperature is decreasing, so we should find that $k$ is negative. To solve such an equation, we may define the function $S(t)=T(t)-3$, which has the same derivative as $T(t)$ and therefore satisfies the following simpler equation $S'(t)=kS(t).$ The solution to the above is $S(t)=Ce^{kt},$ for some constant $C$. We have then $T(t)=Ce^{kt}+3.$ The constants are determined by the other data of the question, which is the initial condition together with the data of the temperature at $t=10$. The temperature of the apartment is the initial temperature of the water: $T(0)=23$. Thus $23=Ce^{k\cdot 0}+3=C+3,$ from which we conclude that $C=20$. The other information in the question is $T(10)=13$. Thus $13=20e^{10k}+3\Rightarrow e^{10k}={\frac {1}{2}}\Rightarrow k={\frac {1}{10}}\cdot \ln({\frac {1}{2}})={\frac {1}{10}}\cdot \ln(2^{-1})=-{\frac {1}{10}}\cdot \ln(2),$ which is consistent with our comment above about $k<0$. We may now write down the solution: $\color {blue}T(t)=20e^{\ln(2)\cdot {\frac {-t}{10}}}+3=20\left(e^{\ln(2)}\right)^{\frac {-t}{10}}+3=20\cdot 2^{\frac {-t}{10}}+3$