Respuesta :
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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