Explanation Given: Initially the temperature of peach pie, T ( 0 ) = 100 ° C The temperature of peach pie after 10 minutes, T ( 10 ) = 80 ° C The temperature of peach pie after 20 minutes, T ( 20 ) = 65 ° C Result used: Newton’s law of cooling. Rate of change of cooling of an object is proportional to the temperature difference between the object and its surroundings. Rate of cooling is d T d t = − k ( T − T R ) where T is the temperature and T R is the room temperature and , k is proportionality constant. Separate the variables in Newton’s law of cooling, d T ( T − T s ) = − k d t Integrate on both sides, ∫ d T ( T − T s ) = − k ∫ d t ln ( T − T s ) = − k t + C Take exponential on both sides, e ln ( T − T s ) = e − k t + C T − T R = e − k t e C T − T R = C e − k t ( ∵ e C = C ) T = T R + C e − k t Thus, T = T R + C e − k t (1) Substitute T ( 0 ) = 100 ° C equation (1), T ( 0 ) = T R + C e − k ( 0 ) 100 = T R + C e 0 ( ∵ e 0 = 1 ) 100 = T R + C T R = 100 − C Thus, T R = 100 − C (2) Substitute T ( 10 ) = 80 ° C equation (1), T ( 10 ) = T R + C e − k ( 10 ) 80 = T R + C e − 10 k Thus, 80 = T R + C e − 10 k (3) Substitute T ( 20 ) = 65 ° C equation (1), T ( 20 ) = T R + C e − k ( 20 ) 65 = T R + C

Single Variable Calculus: Early Transcendentals

8th Edition

ISBN: 9781305270336

Author: James Stewart

Publisher: Cengage Learning

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Chapter 9, Problem 7P

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To find: The room temperature.

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Single Variable Calculus: Early Transcendentals

Ch. 9.1 - Show that y=23ex+e2x is a solution of the...Ch. 9.1 - Prob. 2E Ch. 9.1 - (a) For what values of r does the function y = erx...Ch. 9.1 - (a) For what values of k does the function y = cos...Ch. 9.1 - Which of the following functions are solutions of...Ch. 9.1 - (a) Show that every member of the family of...Ch. 9.1 - Prob. 7E Ch. 9.1 - Prob. 8E Ch. 9.1 - Prob. 9E Ch. 9.1 - Prob. 10E

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The room temperature.

Solution Summary: The author explains Newton's law of cooling, which is proportional to the temperature difference between an object and its surroundings.

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