   Chapter 3.8, Problem 16E

Chapter
Section
Textbook Problem

In a murder investigation, the temperature of the corpse was 32.5°C at 1:30 PM and 30.3°C an hour later. Normal body temperature is 37.0°C and the temperature of the surroundings was 20.0°C. When did the murder take place?

To determine

To find: The time at which the murder take place.

Explanation

Theorem used:

“The only solutions of the differential equation dydt=ky are the exponential functions

y(t)=y(0)ekt.”

Newton’s law of cooling:

The formulate Newton’s law of cooling as a differential equation

dTdt=k(TTs) where k is a constant and Ts is surrounding temperature.

The change of variable y(t)=T(t)Ts

Calculation:

Let T(t) be the temperature of the corpse after t hours.

Given that,

The temperature of the surrounding was 20.0°C. That is, Ts=20.

By newton’s law of cooling,

dTdt=k(T20)

Substitute T20=y and dT=dy in the above equation,

dydt=ky

By theorem as stated above.

y(t)=y(0)ekt (1)

Given that the normal temperature is 20.0°C. That is T(0)=37.

Note that, the change of variable y(t)=T(t)Ts

Substitute t=0 and T(0)=37 in the above equation,

y(0)=T(0)Ts=3720=17

Substitute y(0)=17 in equation (1),

y(t)=17ekt (2)

Given that the body temperature at 1:30 PM was 32.5°C. That is T(t)=32.5.

y(t)=32.520   (QTs=20)=12

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