Science:Math Exam Resources/Courses/MATH100/December 2010/Question 02 (b)

MATH100 December 2010

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Question 02 (b)

Full-Solution Problems. In questions 2-8, justify your answers and show all your work. If a box is provided, write your final answer there. Simplification of answers is not required unless explicitly requested. When an apple is taken from a refrigerator, its temperature is $3^{\circ }C$. After 30 minutes in a $19^{\circ }C$ room, its temperature is $11^{\circ }C$. Find the temperature of the apple 90 minutes after it is taken from the refrigerator, expressed as an integer number of degrees Celsius.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The simplest differential equation we can solve for is ${\begin{aligned}{\frac {dy}{dt}}=ky\end{aligned}}$ which gives the solution ${\begin{aligned}y(t)=y(0)e^{kt}.\end{aligned}}$ How can we make our current equation look like this?

Hint 2
Set ${\begin{aligned}y(t)=T(t)-19\end{aligned}}$ Then use the hint to get the differential equation.

Hint 3
The last step is to solve for k using the given information.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Proceeding as in the hints, let $\displaystyle y(t)=T(t)-19$ Then, Newton's law of cooling reads $\displaystyle {\frac {dy}{dt}}={\frac {dT}{dt}}=k(T(t)-19)=ky(t)$ Solving this differential equation gives $\displaystyle y(t)=y(0)e^{kt}$ In terms of T and using the initial condition $T(0)=3$, we have $\displaystyle T(t)-19=(T(0)-19)e^{kt}=(3-19)e^{kt}=-16e^{kt}$ After 30 minutes, we know $T(30)=11$ and so $\displaystyle -16e^{30k}=T(30)-19=11-19=-8$ Solving for k by isolating and taking logarithms gives $\displaystyle k={\frac {\ln(1/2)}{30}}=-{\frac {\ln(2)}{30}}$ We want to find the temperature after 90 minutes and so ${\begin{aligned}T(90)-19&=-16e^{90\cdot (-\ln(2))/30}\\&=-16e^{-3\ln(2)}\\&=-16e^{\ln(2^{-3})}\\&=-16(1/8)\\&=-2\end{aligned}}$ and thus $T(90)=19-2=17$ degrees celsius

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