Calculating Time of Death Using Newton's Law of Cooling

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SUMMARY

The discussion focuses on calculating the time of death using Newton's Law of Cooling, specifically addressing a scenario where a murder victim's body temperature is recorded at 31°C and then 29°C after one hour, with the surrounding air temperature at 21°C. The calculation for the cooling constant k is derived as k = ln(5/4). Subsequently, the time t is calculated using the formula t = ln(8/5) / ln(4/5), resulting in a negative value for t, which is acceptable as it indicates an earlier time when the body was at a higher temperature of 37°C.

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  • Understanding of Newton's Law of Cooling
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  • Basic knowledge of temperature measurement in Celsius
  • Familiarity with exponential decay equations
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rocomath
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So I have a slight problem since I'm getting a negative for time.

A murder victim is discovered at midnight and the temperature of the body is recorded at 31C. One hour later, the temperature of the body is 29C. Assume that the surrounding air temperature remains constant at 21C.

a) Find k

T=A+(T_0-A)e^{-kt}

T(1)=21+(31-21)e^{-k}=29

k=\ln{\frac 5 4}

b) Solve for t

37=21+(31-21)e^{-t\ln{\frac 5 4}}

16=10e^{\ln{\frac 4 5}^t}

t=\frac{\ln{\frac 8 5}}{\ln{\frac 4 5}}

But this gives me a negative t. Would that be okay since I'm trying to use this t value to compute an earlier time?
 
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Yes.

t=0 when the body is discovered and is at 31 C. So it would have been 37 C at an earlier (negative) time.
 

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