Explanation Given: A glass of milk is moved from a refrigerator with a temperature of 5 ∘ C to a room with a temperature of 20 ∘ C . That is T ( 0 ) = 5 ∘ C , A = 20 ∘ C , t = 1 ∘ C and T ( t ) = 7 ∘ C . Recall: The general solution of cooling equation is T ( t ) = C e − k t + A . The solution of the initial value problem for Newton’s law of cooling is T ( t ) = ( T 0 − A ) e − k t + A . Calculation: Substitute the value T ( 0 ) = 5 ∘ C , A = 20 ∘ C , T ( t ) = 7 ∘ C and t = 1 ∘ C in the initial value problem for Newton’s law of cooling equation T ( t ) = ( T 0 − A ) e − k t + A . That is, 7 = ( 5 − 20 ) e − k + 20 ( 5 − 20 ) e − k = 7 − 20 − 15 e − k = − 13 e − k = 13 15 Take natural logarithm on both sides as follows, ln <

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ISBN: 9780136880677

Author: William Briggs

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Chapter 9.4, Problem 29E

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The time taken by the milk having a temperature such that it has 90% of the ambient temperature.

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Chapter 9.4, Problem 28E

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The time taken by the milk having a temperature such that it has 90 % of the ambient temperature.

Solution Summary: The author calculates the time taken by the milk having a temperature such that it has 90% of the ambient temperature.

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The time taken by the milk having a temperature such that it has 90% of the ambient temperature.

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