A dead body was found within a closed room of a house where the temperature was a constant 70° F. At the time of discovery the core temperature of the body was determined to be 80° F. One hour later a second measurement showed that the core temperature of the body was 75° F. Assume that the time of death corresponds to t = 0 and that the core temperature at that time was 98.6° F. Determine how many hours elapsed before the body was found. [Hint: Let t1 > 0 denote the time that the body was discovered.] (Round your answer to one decimal place.)

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Answer:

1.6 hours

Step-by-step explanation:

I started off with T(t)=70+Ce^kt

then since the initial temp was 98.6 I did T(0)=98.6=70+C so C=28.6

Then T(1) = 80 = 28.6e^k + 70

k = ln (10/28.6)

Then plugged that into

T(t)=85=28.6e^ln(10/28.6)t + 70

and got t=.61

The answer says it is about 1.6 hours.

The time that has elapsed before the body was found is 1.5 hour

The given parameters;

  • when time, t = 0, temperature = 98.6 ⁰F
  • when time, t = 1, temperature = 75⁰
  • when time,  t = t, temperature = 80⁰

Apply the Newton's method of cooling equation;

[tex]T(t) = T_{s} + (T_{o} - T_{s})e^{kt}\\\\T(t) = 70 + (98.6 - 70)e^{kt}\\\\T(t) = 70 + 28.6e^{kt}[/tex]

At the time of discovery, we have the following equation,

[tex]T_{t} = 70 + 28.6e^{kt}\\\\80 = 70 + 28.6e^{kt}\\\\10 = 28.6k^{kt}[/tex]

1 hour later, t + 1, we have the second equation;

[tex]75 = 70 + 28.6e^{kt} \\\\5 = 28.6e^{k(t+ 1)} \\\\5 = 28.6e^{kt + k} ---- (2)[/tex]

divide equation 1  by equation 2;

[tex]\frac{10}{5} = \frac{28.6e^{kt}}{28.6 e^{kt + k}} \\\\2 = e^{kt - kt - k}\\\\2 = e^{-k}\\\\-k = ln(2)\\\\k = -0.693[/tex]

The time when he dead body was discovered is calculated as;

[tex]10 = 28.6e^{kt}\\\\10= 28.6e^{-0.693t}\\\\e^{-0.693t} = \frac{10}{28.6} \\\\-0.693 t = ln(\frac{10}{28.6} )\\\\-0.693t = -1.05\\\\t = \frac{1.05}{0.693} \\\\t = 1.515 \ \\\\t \approx 1.5 \ hr[/tex]

Thus, the time that has elapsed before the body was found is 1.5 hour

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