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?

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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
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.