Explanation Given information: Object A and B are placed in a warm bath at temperature T 0 = 40 ° C . Object A has initial temperature − 20 ° C and cooling constant k = 0.004 s − 1 . Object B has initial temperature 0 o C and cooling constant k = 0.002 s − 1 . Formula used: Newton’s law of cooling: y ' = − k ( y − T 0 ) , y ( t ) is the temperature of object, T 0 is ambient temperature and k is cooling constant. The solution to the equation of form y ' = − k ( y − T 0 ) is y ( t ) = T 0 + C e k t , Calculation: A : y 1 ' = − k 1 ( y 1 − T 0 ) B : y 2 ' = − k 2 ( y 2 − T 0 ) From given information, we have T 0 = 40 ° C y 1 ( 0 ) = − 20 ° C k 1 = 0.004 s − 1 y 2 ( 0 ) = 0 ° C k 2 = 0.002 s − 1 A : y 1 ' = − 0.004 ( y 1 − 40 ) The solution is y 1 ( t ) = 40 + C e 0.004 t S i n c e y 1 ( 0 ) = − 20 ° C ⇒ 40 + C e 0 = − 20 ⇒ C = − 20 − 40 ⇒ C = − 60 The solution becomes y 1 ( t ) = 40 − 60 e 0

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Differential Equations

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Chapter 10.2, Problem 11E

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To plot:The temperatures of A and B for 0≤t≤1000 and to find the time after that the objects has the same temperature.

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Chapter 10.2, Problem 10E

Chapter 10.2, Problem 12E

Chapter 10 Solutions

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To plot: The temperatures of A and B for 0 ≤ t ≤ 1000 and to find the time after that the objects has the same temperature.

Solution Summary: The author plots the temperatures of A and B for a long run of t and to find the time after that the objects have the same temperature.

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To plot:The temperatures of A and B for 0≤t≤1000 and to find the time after that the objects has the same temperature.

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Chapter 10 Solutions