Explanation Using the given parameters and using Newton’s Law of Cooling, we determine the values of the constants and finally arrive at the time of death. Given: At 10 : 30 A M , detectives discover a dead body in a room and measure its temperature at 26 ℃. One hour later, its temperature had dropped to 24.8 ℃. Calculation: Let t = 0 be the time when the person died, and let t 0 denote the time, 10 : 30 A M . The differential equation satisfied by the body temperature y ( t ) is y ' = − k ( y − 20 ) by Newton’s Law of Cooling. Separating variables gives d y ( y − 20 ) = − k d t Integrating on both sides, we get ∫ d y ( y − 20 ) = − k ∫ d t = > y = 20 + A e − k t where A = ± e C is a constant. Since the normal body temperature is 37 ℃ we have y ( 0 ) = 37 = 20 + A = > A = 17 To determine k , substitute this value in the given equation: 26 = 20 + 17 e − k t 0 and 24...

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Chapter 10.2, Problem 7E

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The time of death assuming the normal body temperature was $37$ ℃ and the temperature in the room was held constant at $20$ ℃.

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Chapter 10.2, Problem 6E

Chapter 10.2, Problem 8E

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The time of death assuming the normal body temperature was 37 ℃ and the temperature in the room was held constant at 20 ℃.

Solution Summary: The author calculates the time of death using Newton's Law of Cooling and the given parameters.

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The time of death assuming the normal body temperature was $37$ ℃ and the temperature in the room was held constant at $20$ ℃.

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Chapter 10 Solutions

The time of death assuming the normal body temperature was 37 ℃ and the temperature in the room was held constant at 20 ℃.

Solution Summary: The author calculates the time of death using Newton's Law of Cooling and the given parameters.

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The time of death assuming the normal body temperature was 37 ℃ and the temperature in the room was held constant at 20℃.

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Chapter 10 Solutions

The time of death assuming the normal body temperature was $37$ ℃ and the temperature in the room was held constant at $20$ ℃.