Science:Math Exam Resources/Courses/MATH100/December 2018/Question 09

MATH100 December 2018

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Question 09
A can of soda was placed in the refrigerator which operates at the constant temperature of $5$ degrees Celsius. Originally, the can has temperature $20$ degrees Celsius, but then at 1:00 p.m., the temperature of soda can is $15$ degrees Celsius, while at 1:45 p.m., the temperature of the soda can is $10$ degrees Celsius. At what time was the soda can put 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 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
Recall that Newton's law of cooling is ${\frac {{\rm {d}}T}{{\rm {d}}t}}=k(T-A)$ , where $T=T(t)$ is the body's temperature at time $t$ , $A$ is the ambient temperature, and $k<0$ is a constant.

Hint 2
To solve a differential equation of the form ${\frac {{\rm {d}}T}{{\rm {d}}t}}=k(T-A)$ , we can make the substitution $f(t)=T(t)-A$ .

Hint 3
It is convenient to measure the time $t$ relative to 1 p.m. Thus, for instance, $T(0)=15$ (where $T$ is the temperature in degrees Celsius of the can).

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Solution
We denote by $T(t)$ the temperature function (with respect to time $t$ , which is measured in minutes, while the temperature is measured in degrees Celsius) for the soda can. The initial time $t=0$ is when we made the first reading of the temperature of the soda can (i.e., 1 p.m.). So, we know that $T(0)=15$ and $T(45)=10$ . According to Newton's Law of Cooling, we have the following equation: ${\frac {{\rm {d}}T}{{\rm {d}}t}}=k(T-5)$ for some negative constant $k$ . We let $f(t)=T(t)-5$ ; then $f'(t)=T'(t)$ and therefore, we have $f'(t)=kf(t)$ , which means that $f(t)=f(0)e^{kt}$ , where $f(0)=T(0)-5=15-5=10$ . So, $f(t)=10e^{kt}$ and therefore, $T(t)=5+10e^{kt}$ . Because $T(45)=10$ , we get that $5+10e^{k\,\cdot \,45}=10$ and so, $10e^{45k}=5$ , which means that $e^{45k}={\frac {1}{2}}$ and so, $k=-{\frac {\log(2)}{45}}$ . Thus, $T(t)=5+10e^{-{\frac {\log(2)\cdot t}{45}}}$ and we need to find the time $t_{0}$ when the can was put in the refrigerator, i.e., when $T(t_{0})=20$ ; note that $t_{0}$ will be negative since the soda can was placed in the refrigerator prior to time $t=0$ , represented by 1 p.m. So, we have $T(t_{0})=5+10e^{-{\frac {\log(2)\cdot t_{0}}{45}}}=20$ and thus $10e^{-{\frac {\log(2)t_{0}}{45}}}=15$ , which yields $e^{-{\frac {\log(2)t_{0}}{45}}}=1.5$ and so, $-{\frac {\log(2)t_{0}}{45}}=\log(1.5)$ and so, $t_{0}=-{\frac {45\log(1.5)}{\log(2)}}$ . Therefore, the soda can was placed in the refrigerator $\color {blue}{\frac {45\log(1.5)}{\log(2)}}$ minutes prior to 1 p.m.