Find out the answer to "Applications of First Order Differential Equations: A dead..."

Carla Murphy

Answered question

2022-01-15

Applications of First Order Differential Equations: A dead body was found within a closed room of a house where the temperature was a constant

24^{\circ} C

. At the time of discovery, the core temperature of the body was determined to be

28^{\circ} C

. One hour later, a second measurement showed that the core temperature of the body was

26^{\circ} C

. The core temperature of the body at the time of death is

37^{\circ} C

. A. Determine how many hours elapsed before the body was found. B. What is the temperature of the body 5 hours from the death of the victim?

Answer & Explanation

Bernard Lacey

Beginner2022-01-16Added 30 answers

From Newtons

GaceCoect5v

Beginner2022-01-17Added 26 answers

$T^{'} = k (T - T_{m})$ (Newton's Law of Cooling)
$\int \frac{d T}{T - T_{m}} = \int k d t$
$\ln | T - T_{m} | = k t + C_{1} \to T (t) = T_{m} + c e^{k t}$
$T (closed room) = 21 \to T_{m} = 21$
$T (temperature at t = 0) = 37 \to T (0) = 37$
Substitute in $T (t) = T_{m} + c e^{k t}$
$37 = 21 + c e^{0} \to c = 16$
$T = 21 + 16 e^{k t}$ (1)
where t denotes the discovery time
The temperature was $27^{\circ}$ at the time of discovery
$27 = 21 + 16 e^{k t} \to 16 e^{k t} = 6$ (2)
The temperature was $25^{\circ}$ after one hour $(t + 1)$
$25 = 21 + 16 e^{k (t + 1)} \to 16 e^{k (t + 1)} = 4$ (3)
Dividing (2) by (3)
$\frac{e^{k t}}{e^{k (t + 1)}} = \frac{6}{4} = \frac{3}{2}$
$e^{k t - k t - k} = \frac{3}{2} \to e^{- k} = \frac{3}{2} \to - k = \ln \frac{3}{2} \to k = - 0.41$
Substitute in (2)
$16 e^{- 0.41 t} = 6 \to e^{- 0.41 t} = \frac{6}{16} = \frac{3}{8}$
$- 0.41 t = \ln \frac{3}{8}$
$t = \ln \frac{3}{8} \div (- 0.41)$
$t = 2.4$ hours

alenahelenash

Expert2022-01-24Added 556 answers

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