Theorem 6.5.1. Newton's Law of Cooling. The temperature T at a time x of a cooling object follows the function T=A+Be^kxA is the ambient (or room) temperature. B and k are constants that depend on the object.Suppose a forensics technician arrived at a murder scene and recorded the temperature of the surroundings as well as the body. The technician decides it is fair to assume that the room temperature has been holding steady at about 68 °F. A thermometer was placed in the liver of the corpse and the following table of values was recorded.Table 6.5.2. Time and TemperatureActual TimeMinutes Elapsed (x)Temperature, ,T, of the Body (°F)2:00 pm085.902:20 pm2085.172:40 pm4084.473:00 pm6083.78The key to estimating time of death is to estimate ,A, ,B, ,k, and x in First question: Recall that the technician thinks the room temp was 68 °F. By substituting this number and the first recording (0, 85.9) into the Cooling Equation, find:A:BSecond Question: Once you know A and ,B, substitute some other data point into the equation so that k is the only variable. Solve the resulting equation and round k to 6 decimal places. Show your work. Remember the exponential must be isolated before you take the natural log of both sides.Find: k≈ Question three: Note: the number k is called the cooling (or warming) constant. If an object cools, k should be negative. Mathematically, looking at the equation, why should k be negative? Theorem 6.5.1. Newton's Law of Cooling. The temperature T at a timex of a cooling object follows the functionT = A+ Bek.||A is the ambient (or room) temperature. B and k are constants that depend onthe object.Suppose a forensics technician arrived at a murder scene and recorded thetemperature of the surroundings as well as the body. The technician decides itis fair to assume that the room temperature has been holding steady at about68 °F. A thermometer was placed in the liver of the corpse and the followingtable of values was recorded.Table 6.5.2. Time and TemperatureActual Time Minutes Elapsed (x)|Temperature, T, of the Body (°F)2:00 pm85.902:20 pm2085.172:40 pm4084.473:00 pm6083.78The key to estimating time of death is to estimate A, B, k, and x inTheorem 6.5.1. (a) Recall that the technician thinks the room temp was 68 °F. By substitutingthis number and the first recording (0, 85.9) into the Cooling Equation, findAand B =(b) Once you know A and B, substitute some other data point into theequation so that k is the only variable. Solve the resulting equation and round kto 6 decimal places. Show your work. Remember the exponential must beisolated before you take the natural log of both sides.(c) Note: the number k is called the cooling (or warming) constant. If an objectcools, k should be negative. Mathematically, looking at the equation, whyshould k be negative?

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The temperature T at a time x of a cooling object follows the function T=A+Be^kx A is the ambient (or room) temperature. B and k are constants that depend on the object. Suppose a forensics technician arrived at a murder scene and recorded the temperature of the surroundings as well as the body. The technician decides it is fair to assume that the room temperature has been holding steady at about 68 °F. A thermometer was placed in the liver of the corpse and the following table of values was recorded. Table 6.5.2. Time and Temperature Actual Time Minutes Elapsed (x) Temperature, ,T, of the Body (°F) 2:00 pm 0 85.90 2:20 pm 20 85.17 2:40 pm 40 84.47 3:00 pm 60 83.78 The key to estimating time of death is to estimate ,A, ,B, ,k, and x in First question: Recall that the technician thinks the room temp was 68 °F. By substituting this number and the first recording (0, 85.9) into the Cooling Equation, find: A: B Second Question: Once you know A and ,B, substitute some other data point into the equation so that k is the only variable. Solve the resulting equation and round k to 6 decimal places. Show your work. Remember the exponential must be isolated before you take the natural log of both sides. Find: k≈ Question three: Note: the number k is called the cooling (or warming) constant. If an object cools, k should be negative. Mathematically, looking at the equation, why should k be negative?

Elements Of Electromagnetics

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ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

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Problem 1.1MA

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Theorem 6.5.1. Newton's Law of Cooling.

The temperature T at a time x of a cooling object follows the function ^{T=A+Be^}^kx

A is the ambient (or room) temperature. B and k are constants that depend on the object.

Suppose a forensics technician arrived at a murder scene and recorded the temperature of the surroundings as well as the body. The technician decides it is fair to assume that the room temperature has been holding steady at about 68 °F. A thermometer was placed in the liver of the corpse and the following table of values was recorded.

Table 6.5.2. Time and Temperature

Actual Time	Minutes Elapsed (x)	Temperature, ,T, of the Body (°F)
2:00 pm	0	85.90
2:20 pm	20	85.17
2:40 pm	40	84.47
3:00 pm	60	83.78

The key to estimating time of death is to estimate ,A, ,B, ,k, and x in

First question: Recall that the technician thinks the room temp was 68 °F. By substituting this number and the first recording (0, 85.9) into the Cooling Equation, find:

A:

B

Second Question: Once you know A and ,B, substitute some other data point into the equation so that k is the only variable. Solve the resulting equation and round k to 6 decimal places. Show your work. Remember the exponential must be isolated before you take the natural log of both sides.

Find: k≈

Question three: Note: the number k is called the cooling (or warming) constant. If an object cools, k should be negative. Mathematically, looking at the equation, why should k be negative?

Theorem 6.5.1. Newton's Law of Cooling. The temperature T at a time
x of a cooling object follows the function
T = A+ Bek.
||
A is the ambient (or room) temperature. B and k are constants that depend on
the object.
Suppose a forensics technician arrived at a murder scene and recorded the
temperature of the surroundings as well as the body. The technician decides it
is fair to assume that the room temperature has been holding steady at about
68 °F. A thermometer was placed in the liver of the corpse and the following
table of values was recorded.
Table 6.5.2. Time and Temperature
Actual Time Minutes Elapsed (x)|Temperature, T, of the Body (°F)
2:00 pm
85.90
2:20 pm
20
85.17
2:40 pm
40
84.47
3:00 pm
60
83.78
The key to estimating time of death is to estimate A, B, k, and x in
Theorem 6.5.1.

(a) Recall that the technician thinks the room temp was 68 °F. By substituting
this number and the first recording (0, 85.9) into the Cooling Equation, find
A
and B =
(b) Once you know A and B, substitute some other data point into the
equation so that k is the only variable. Solve the resulting equation and round k
to 6 decimal places. Show your work. Remember the exponential must be
isolated before you take the natural log of both sides.
(c) Note: the number k is called the cooling (or warming) constant. If an object
cools, k should be negative. Mathematically, looking at the equation, why
should k be negative?

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