Newton's Law of Cooling states that a hot object cools down at a rate proportional to the temperature difference between the object and its surroundings. For example, consider a pot of hot water a temperature of 90°C cooling down in a room that is kept at 20°C. Its temperature (7) after cooling for t minutes is given by: T(t) = 20+70e-0.034 / t>0 " (a) Show that this function gives the object's correct initial temperature. (b) What is the object's temperature after cooling for 10 minutes (to the nearest tenth of a degree)? (c) At what rate (in °C/min) is the object cooling after 20 minutes? (d) Using the original function, determine what the object's temperature will become (after a long period of time). Does this result match with what you would expect? Explain.
Newton's Law of Cooling states that a hot object cools down at a rate proportional to the temperature difference between the object and its surroundings. For example, consider a pot of hot water a temperature of 90°C cooling down in a room that is kept at 20°C. Its temperature (7) after cooling for t minutes is given by: T(t) = 20+70e-0.034 / t>0 " (a) Show that this function gives the object's correct initial temperature. (b) What is the object's temperature after cooling for 10 minutes (to the nearest tenth of a degree)? (c) At what rate (in °C/min) is the object cooling after 20 minutes? (d) Using the original function, determine what the object's temperature will become (after a long period of time). Does this result match with what you would expect? Explain.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Newton's Law of Cooling states that a hot object cools down at a rate
proportional to the temperature difference between the object and its
surroundings. For example, consider a pot of hot water a temperature of 90°C
cooling down in a room that is kept at 20°C. Its temperature (7) after cooling for t
minutes is given by:
T(t) = 20+70e-0.0341 t≥0
(a)
Show that this function gives the object's correct initial temperature.
(b)
What is the object's temperature after cooling for 10 minutes (to the
nearest tenth of a degree)?
At what rate (in °C/min) is the object cooling after 20 minutes?
Using the original function, determine what the object's temperature will
become (after a long period of time). Does this result match with what you
would expect? Explain.
(c)
(d)
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