Science:Math Exam Resources/Courses/MATH102/December 2017/Question 08

MATH102 December 2017

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Question 08

A freshly laid egg has a temperature of $40$ degrees, and the temperature outside is $10$ degrees. After ten minutes ($t=10$), the temperature of the egg is $25$ degrees. What will the temperature be after $10$ more minutes ($t=20$)? Assume the system obeys Newton’s Law of Cooling. That is, the rate of change of the temperature of the egg is proportional to the difference in temperature between the egg and the outside. Justify your answer. Hint: you can find an exact answer with minimal calculation (and, of course, no calculator).

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
First write down an exponential relation involving unknown constants. Then solve for the 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 us denote by ${\textstyle T(t)}$ the temperature of the egg at time ${\textstyle t}$. By the Newton’s Law of Cooling, $$(T(t)-10)'=-k(T(t)-10).$$ for some constant ${\textstyle k}$. The solution of this differential equation can be written as $$T(t)-10=Ce^{-kt},$$ where ${\textstyle C}$ is some constant. According to the description, we know that ${\textstyle T(0)=40}$ and ${\textstyle T(10)=25}$. So $$40-10=Ce^{0},$$ $$25-10=Ce^{-10k}.$$ The first equation implies that ${\textstyle C=30}$. Plugging into the second one, $$15=30e^{-10k},$$ $$e^{-10k}={\dfrac {1}{2}}.$$ Now, at time ${\textstyle t=20}$, we have $$T(20)-10=30e^{-20k}.$$ Using the rule of exponents, we have $$e^{-20k}=(e^{-10k})^{2}=\left({\dfrac {1}{2}}\right)^{2}={\dfrac {1}{4}}.$$ So $$T(20)-10=30\cdot {\dfrac {1}{4}},$$ $$T(20)=10+{\dfrac {30}{4}}=10+{\dfrac {15}{2}}={\dfrac {35}{2}}=17.5.$$ Answer: the temperature after ${\textstyle 10}$ more minutes is ${\textstyle {\color {blue}17.5}}$ degrees.