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2. (Separable and Newton's Law of Cooling) Let T(t) be the temperature of the room and T, the outside temperature (assumed to be constant in Newton's law). Then by Newton's law k(T - T,). In this problem, the rate at which a substance cools in air is directly proportional to the difference between the temperatures of the substance and that of air. The temperature of air, TA, is 30° and the substance cools from 100' to 70' in 15 minutes. How long does it take for the substance to cool from 100' to 50'? A. 45.30 minutes B. 43.60 minutes C. 35.39 minutes D. 33.59 minutes

2. (Separable and Newton's Law of Cooling) Let T(t) be the temperature of the room and T, the outside temperature (assumed to be constant in Newton's law). Then by Newton's law k(T - T,). In this problem, the rate at which a substance cools in air is directly proportional to the difference between the temperatures of the substance and that of air. The temperature of air, TA, is 30° and the substance cools from 100' to 70' in 15 minutes. How long does it take for the substance to cool from 100' to 50'? A. 45.30 minutes B. 43.60 minutes C. 35.39 minutes D. 33.59 minutes

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section: Chapter Questions
Problem 18T
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2. (Separable and Newton's Law of Cooling) Let T(t) be the temperature of the room and T, the outside temperature (assumed to be constant in Newton's law). Then by Newton's law dt k(T – T,). In this problem, the rate at which a substance cools in air is directly proportional to the difference between the temperatures of the substance and that of air. The temperature of air, TA, is 30° and the substance cools from 100° to 70° in 15 minutes. How long does it take for the substance to cool from 100 to 50"? A. 45.30 minutes B. 43.60 minutes C. 35.39 minutes D. 33.59 minutes Solution. Hints: 1. Find the general solution of the equation = k(T – T,) by separable. 2. Let t = 0 and T = 100° to find C. 3. Let t = 15 and T = 70 to find k. 4. Let T = 50° to find t.
Transcribed Image Text:2. (Separable and Newton's Law of Cooling) Let T(t) be the temperature of the room and T, the outside temperature (assumed to be constant in Newton's law). Then by Newton's law dt k(T – T,). In this problem, the rate at which a substance cools in air is directly proportional to the difference between the temperatures of the substance and that of air. The temperature of air, TA, is 30° and the substance cools from 100° to 70° in 15 minutes. How long does it take for the substance to cool from 100 to 50"? A. 45.30 minutes B. 43.60 minutes C. 35.39 minutes D. 33.59 minutes Solution. Hints: 1. Find the general solution of the equation = k(T – T,) by separable. 2. Let t = 0 and T = 100° to find C. 3. Let t = 15 and T = 70 to find k. 4. Let T = 50° to find t.
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