Explanation Use the general equation of Newton’s Law of Cooling to obtain the value of time required to cool an object from temperature A to temperature B to verify the given statement Given: t = 1 k ln ( A − T 0 B − T 0 ) where T 0 is the ambient temperature Calculation: At a given time t , the temperature of the object is y ( t ) = T 0 + C e − k t for the constant C . Suppose the object is initially at a temperature A and reaches temperature B at time t , then A = T 0 + C = > C = A − T 0 Also, B = T 0 +

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

See similar textbooks

Videos

Question

Chapter 10.2, Problem 25E

To determine

To show that by Newton’s law of Cooling, the time required to cool an object from temperature $A$ to temperature $B$ is $t = \frac{1}{k} \ln (\frac{A - T_{0}}{B - T_{0}})$ where $T_{0}$ is the ambient temperature

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10.2, Problem 24E

Chapter 10.2, Problem 26E

Students have asked these similar questions

Show that i cote +1 = cosec 20 tan 20+1 = sec² O २ cos² + sin 20 = 1 using pythagon's theorem

Find the general solution to the differential equation

charity savings Budget for May travel food Peter earned $700 during May. The graph shows how the money was used. What fraction was clothes? O Search Submit clothes leisure

Chapter 10 Solutions

CALCULUS (CLOTH)

Ch. 10.1 - Prob. 1PQ Ch. 10.1 - Prob. 2PQ Ch. 10.1 - Prob. 3PQ Ch. 10.1 - Prob. 4PQ Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E

Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E Ch. 10.1 - Prob. 11E Ch. 10.1 - Prob. 12E Ch. 10.1 - Prob. 13E Ch. 10.1 - Prob. 14E Ch. 10.1 - Prob. 15E Ch. 10.1 - Prob. 16E Ch. 10.1 - Prob. 17E Ch. 10.1 - Prob. 18E Ch. 10.1 - Prob. 19E Ch. 10.1 - Prob. 20E Ch. 10.1 - Prob. 21E Ch. 10.1 - Prob. 22E Ch. 10.1 - Prob. 23E Ch. 10.1 - Prob. 24E Ch. 10.1 - Prob. 25E Ch. 10.1 - Prob. 26E Ch. 10.1 - Prob. 27E Ch. 10.1 - Prob. 28E Ch. 10.1 - Prob. 29E Ch. 10.1 - Prob. 30E Ch. 10.1 - Prob. 31E Ch. 10.1 - Prob. 32E Ch. 10.1 - Prob. 33E Ch. 10.1 - Prob. 34E Ch. 10.1 - Prob. 35E Ch. 10.1 - Prob. 36E Ch. 10.1 - Prob. 37E Ch. 10.1 - Prob. 38E Ch. 10.1 - Prob. 39E Ch. 10.1 - Prob. 40E Ch. 10.1 - Prob. 41E Ch. 10.1 - Prob. 42E Ch. 10.1 - Prob. 43E Ch. 10.1 - Prob. 44E Ch. 10.1 - Prob. 45E Ch. 10.1 - Prob. 46E Ch. 10.1 - Prob. 47E Ch. 10.1 - Prob. 48E Ch. 10.1 - Prob. 49E Ch. 10.1 - Prob. 50E Ch. 10.1 - Prob. 51E Ch. 10.1 - Prob. 52E Ch. 10.1 - Prob. 53E Ch. 10.1 - Prob. 54E Ch. 10.1 - Prob. 55E Ch. 10.1 - Prob. 56E Ch. 10.1 - Prob. 57E Ch. 10.1 - Prob. 58E Ch. 10.1 - Prob. 59E Ch. 10.1 - Prob. 60E Ch. 10.1 - Prob. 61E Ch. 10.1 - Prob. 62E Ch. 10.1 - Prob. 63E Ch. 10.1 - Prob. 64E Ch. 10.1 - Prob. 65E Ch. 10.1 - Prob. 66E Ch. 10.1 - Prob. 67E Ch. 10.1 - Prob. 68E Ch. 10.1 - Prob. 69E Ch. 10.1 - Prob. 70E Ch. 10.1 - Prob. 71E Ch. 10.2 - Prob. 1PQ Ch. 10.2 - Prob. 2PQ Ch. 10.2 - Prob. 3PQ Ch. 10.2 - Prob. 4PQ Ch. 10.2 - Prob. 1E Ch. 10.2 - Prob. 2E Ch. 10.2 - Prob. 3E Ch. 10.2 - Prob. 4E Ch. 10.2 - Prob. 5E Ch. 10.2 - Prob. 6E Ch. 10.2 - Prob. 7E Ch. 10.2 - Prob. 8E Ch. 10.2 - Prob. 9E Ch. 10.2 - Prob. 10E Ch. 10.2 - Prob. 11E Ch. 10.2 - Prob. 12E Ch. 10.2 - Prob. 13E Ch. 10.2 - Prob. 14E Ch. 10.2 - Prob. 15E Ch. 10.2 - Prob. 16E Ch. 10.2 - Prob. 17E Ch. 10.2 - Prob. 18E Ch. 10.2 - Prob. 19E Ch. 10.2 - Prob. 20E Ch. 10.2 - Prob. 21E Ch. 10.2 - Prob. 22E Ch. 10.2 - Prob. 23E Ch. 10.2 - Prob. 24E Ch. 10.2 - Prob. 25E Ch. 10.2 - Prob. 26E Ch. 10.3 - Prob. 1PQ Ch. 10.3 - Prob. 2PQ Ch. 10.3 - Prob. 3PQ Ch. 10.3 - Prob. 4PQ Ch. 10.3 - Prob. 1E Ch. 10.3 - Prob. 2E Ch. 10.3 - Prob. 3E Ch. 10.3 - Prob. 4E Ch. 10.3 - Prob. 5E Ch. 10.3 - Prob. 6E Ch. 10.3 - Prob. 7E Ch. 10.3 - Prob. 8E Ch. 10.3 - Prob. 9E Ch. 10.3 - Prob. 10E Ch. 10.3 - Prob. 11E Ch. 10.3 - Prob. 12E Ch. 10.3 - Prob. 13E Ch. 10.3 - Prob. 14E Ch. 10.3 - Prob. 15E Ch. 10.3 - Prob. 16E Ch. 10.3 - Prob. 17E Ch. 10.3 - Prob. 18E Ch. 10.3 - Prob. 19E Ch. 10.3 - Prob. 20E Ch. 10.3 - Prob. 21E Ch. 10.3 - Prob. 22E Ch. 10.3 - Prob. 23E Ch. 10.3 - Prob. 24E Ch. 10.3 - Prob. 25E Ch. 10.3 - Prob. 26E Ch. 10.3 - Prob. 27E Ch. 10.3 - Prob. 28E Ch. 10.3 - Prob. 29E Ch. 10.4 - Prob. 1PQ Ch. 10.4 - Prob. 2PQ Ch. 10.4 - Prob. 3PQ Ch. 10.4 - Prob. 1E Ch. 10.4 - Prob. 2E Ch. 10.4 - Prob. 3E Ch. 10.4 - Prob. 4E Ch. 10.4 - Prob. 5E Ch. 10.4 - Prob. 6E Ch. 10.4 - Prob. 7E Ch. 10.4 - Prob. 8E Ch. 10.4 - Prob. 9E Ch. 10.4 - Prob. 10E Ch. 10.4 - Prob. 11E Ch. 10.4 - Prob. 12E Ch. 10.4 - Prob. 13E Ch. 10.4 - Prob. 14E Ch. 10.4 - Prob. 15E Ch. 10.4 - Prob. 16E Ch. 10.4 - Prob. 17E Ch. 10.4 - Prob. 18E Ch. 10.5 - Prob. 1PQ Ch. 10.5 - Prob. 2PQ Ch. 10.5 - Prob. 3PQ Ch. 10.5 - Prob. 4PQ Ch. 10.5 - Prob. 1E Ch. 10.5 - Prob. 2E Ch. 10.5 - Prob. 3E Ch. 10.5 - Prob. 4E Ch. 10.5 - Prob. 5E Ch. 10.5 - Prob. 6E Ch. 10.5 - Prob. 7E Ch. 10.5 - Prob. 8E Ch. 10.5 - Prob. 9E Ch. 10.5 - Prob. 10E Ch. 10.5 - Prob. 11E Ch. 10.5 - Prob. 12E Ch. 10.5 - Prob. 13E Ch. 10.5 - Prob. 14E Ch. 10.5 - Prob. 15E Ch. 10.5 - Prob. 16E Ch. 10.5 - Prob. 17E Ch. 10.5 - Prob. 18E Ch. 10.5 - Prob. 19E Ch. 10.5 - Prob. 20E Ch. 10.5 - Prob. 21E Ch. 10.5 - Prob. 22E Ch. 10.5 - Prob. 23E Ch. 10.5 - Prob. 24E Ch. 10.5 - Prob. 25E Ch. 10.5 - Prob. 26E Ch. 10.5 - Prob. 27E Ch. 10.5 - Prob. 28E Ch. 10.5 - Prob. 29E Ch. 10.5 - Prob. 30E Ch. 10.5 - Prob. 31E Ch. 10.5 - Prob. 32E Ch. 10.5 - Prob. 33E Ch. 10.5 - Prob. 34E Ch. 10.5 - Prob. 35E Ch. 10.5 - Prob. 36E Ch. 10.5 - Prob. 37E Ch. 10.5 - Prob. 38E Ch. 10.5 - Prob. 39E Ch. 10.5 - Prob. 40E Ch. 10.5 - Prob. 41E Ch. 10.5 - Prob. 42E Ch. 10.5 - Prob. 43E Ch. 10.5 - Prob. 44E Ch. 10.5 - Prob. 45E Ch. 10.5 - Prob. 46E Ch. 10.5 - Prob. 47E Ch. 10.5 - Prob. 48E Ch. 10.5 - Prob. 49E Ch. 10 - Prob. 1CRE Ch. 10 - Prob. 2CRE Ch. 10 - Prob. 3CRE Ch. 10 - Prob. 4CRE Ch. 10 - Prob. 5CRE Ch. 10 - Prob. 6CRE Ch. 10 - Prob. 7CRE Ch. 10 - Prob. 8CRE Ch. 10 - Prob. 9CRE Ch. 10 - Prob. 10CRE Ch. 10 - Prob. 11CRE Ch. 10 - Prob. 12CRE Ch. 10 - Prob. 13CRE Ch. 10 - Prob. 14CRE Ch. 10 - Prob. 15CRE Ch. 10 - Prob. 16CRE Ch. 10 - Prob. 17CRE Ch. 10 - Prob. 18CRE Ch. 10 - Prob. 19CRE Ch. 10 - Prob. 20CRE Ch. 10 - Prob. 21CRE Ch. 10 - Prob. 22CRE Ch. 10 - Prob. 23CRE Ch. 10 - Prob. 24CRE Ch. 10 - Prob. 25CRE Ch. 10 - Prob. 26CRE Ch. 10 - Prob. 27CRE Ch. 10 - Prob. 28CRE Ch. 10 - Prob. 29CRE Ch. 10 - Prob. 30CRE Ch. 10 - Prob. 31CRE Ch. 10 - Prob. 32CRE Ch. 10 - Prob. 33CRE Ch. 10 - Prob. 34CRE Ch. 10 - Prob. 35CRE Ch. 10 - Prob. 36CRE Ch. 10 - Prob. 37CRE Ch. 10 - Prob. 38CRE Ch. 10 - Prob. 39CRE Ch. 10 - Prob. 40CRE Ch. 10 - Prob. 41CRE Ch. 10 - Prob. 42CRE Ch. 10 - Prob. 43CRE Ch. 10 - Prob. 44CRE Ch. 10 - Prob. 45CRE Ch. 10 - Prob. 46CRE Ch. 10 - Prob. 47CRE Ch. 10 - Prob. 48CRE Ch. 10 - Prob. 49CRE Ch. 10 - Prob. 50CRE Ch. 10 - Prob. 51CRE Ch. 10 - Prob. 52CRE

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

To show that by Newton’s law of Cooling, the time required to cool an object from temperature A to temperature B is t = 1 k ln ( A − T 0 B − T 0 ) where T 0 is the ambient temperature

Solution Summary: The author explains how Newton's law of Cooling proves that the time required to cool an object from temperature A to temperature B is 1 k ln.

Videos

To show that by Newton’s law of Cooling, the time required to cool an object from temperature $A$ to temperature $B$ is $t = \frac{1}{k} \ln (\frac{A - T_{0}}{B - T_{0}})$ where $T_{0}$ is the ambient temperature

Want to see the full answer?

Chapter 10 Solutions

To show that by Newton’s law of Cooling, the time required to cool an object from temperature A to temperature B is t = 1 k ln ( A − T 0 B − T 0 ) where T 0 is the ambient temperature

Solution Summary: The author explains how Newton's law of Cooling proves that the time required to cool an object from temperature A to temperature B is 1 k ln.

Videos

To show that by Newton’s law of Cooling, the time required to cool an object from temperature A to temperature B is t=1kln(A−T0B−T0) where T0 is the ambient temperature

Want to see the full answer?

Chapter 10 Solutions

To show that by Newton’s law of Cooling, the time required to cool an object from temperature $A$ to temperature $B$ is $t = \frac{1}{k} \ln (\frac{A - T_{0}}{B - T_{0}})$ where $T_{0}$ is the ambient temperature