Program Explanation Given information: The buttermilk is cooled at 25 ° C initially and the cooling temperature is 0 ° C .The temperature of the buttermilk after 20 minutes is 15 ° C . Calculation: The newton’s law of cooling is the time rate of change of the temperature T ( t ) of an object is proportional to the difference of A − T . Therefore, it can be mathematically expressed as, d T d t = k ( A − T ) Here, k is constant, T is temperature, t is time, A is cooling temperature. Now use the method of separation of the variable to solve the above differential equation. d T d t = k ( T − A ) d T T − A = k d T Now integrate both sides. ∫ d T T − A = ∫ k d T ln T − A = k t + c 1 T = A + c e k t Therefore, the temperature function is as follows, T ( t ) = A + c e k t ........ (1) The given initial condition of cooling of the buttermilk is 25 ° C as T ( 0 ) = 25 ° C . Now apply the initial condition T ( 0 ) = 25 ° C on equation (1) to obtain the value of c . T ( t ) = A + c e k t 25 = 0 + c e k ( 0 ) c = 25 It is given that the temperature of the buttermilk after 20 minutes is 15 ° C as T ( 20 ) = 15 ° C

5th Edition

ISBN: 9780321974235

Author: Calvis

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Chapter 1.4, Problem 43P

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Program Description: Purpose of the problem is to find the time at which the temperature of buttermilk is 5°C .

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Chapter 1.4, Problem 42P

Chapter 1.4, Problem 44P

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Chapter 1 Solutions

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Program Description: Purpose of the problem is to find the time at which the temperature of buttermilk is 5 ° C .

Solution Summary: The author explains how the newton's law of cooling is the time rate of change of the temperature of an object is proportional to the difference of A-T.

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Program Description: Purpose of the problem is to find the time at which the temperature of buttermilk is 5°C .

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Chapter 1 Solutions