Science:Math Exam Resources/Courses/MATH100/December 2011/Question 02 (a)

MATH100 December 2011

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Question 02 (a)

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 stated. A wealthy man was found murdered in his home. The police arrived on scene at 10:00 P.M. The temperature of the body at 10:00 P.M. was 33°C and one hour later it was 31°C. The temperature of the room in which the body was found was 21°C and normal body temperature is 37°C. Assume that the body cools after death according to Newton's Law of Cooling. How many hours before the police arrived on the scene did the murder occur?

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
Recall that Newton's Law of Cooling states that $\displaystyle T(t)=T_{a}+(T_{0}-T_{a})e^{-kt}$ where Ta is the temperature of the environment; T0 is the initial temperature of the object; and T(t) is the temperature of the object at time t.

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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. Newton's Law of Cooling which states that $\displaystyle T(t)=T_{a}+(T_{0}-T_{a})e^{-kt}$. Here, the initial temperature of the body T0 is 33°C; the temperature of the environment Ta is given to be 21°C; and the temperature of the body after 1 hour (T(1)) is 31°C. Plugging in these numbers give us $\displaystyle 31=21+(33-21)e^{-k}$ Using this equation, we can solve for k: ${\begin{aligned}e^{-k}&={\frac {31-21}{33-21}}={\frac {5}{6}}\\-k&=\ln(5/6)\\k&=-\ln(5/6)\\\end{aligned}}$ And so now we have found the equation that gives us the temperature of the body at any time t: $\displaystyle T(t)=21+12e^{t\ln(5/6)}$ and we would like to know the time of death, which is the time at which the body temperature was 37°C. For this, we simply solve $\displaystyle T(t)=37$ that is ${\begin{aligned}21+12e^{t\ln(5/6)}&=37\\12e^{t\ln(5/6)}&=16\\e^{t\ln(5/6)}&=16/12\\t\ln(5/6)&=\ln(4/3)\\t&={\frac {\ln(4/3)}{\ln(5/6)}}\\t&\approx -1.578\end{aligned}}$ Therefore, the police arrived $\ln(4/3)/\ln(5/6)$ hours after the murder, that is, just over an hour and a half.

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