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A cup of hot coffee initially at 95°C cools to 80°C in 6 min while sitting in a room of temperature 21°C. By Newton's law of cooling, the temperature of the hot coffee- is T(t) = M - Ce-Kt. Determine the value of G and K

A cup of hot coffee initially at 95°C cools to 80°C in 6 min while sitting in a room of temperature 21°C. By Newton's law of cooling, the temperature of the hot coffee- is T(t) = M - Ce-Kt. Determine the value of G and K

Advanced Engineering Mathematics
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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A cup of hot coffee initially at 95°C cools to 80°C in 6 min while sitting in a room of temperature 21°C. By Newton's law of cooling, the temperature of the hot coffee is T(t) = M - Ce-kt. Determine the value of C and K.
Transcribed Image Text:A cup of hot coffee initially at 95°C cools to 80°C in 6 min while sitting in a room of temperature 21°C. By Newton's law of cooling, the temperature of the hot coffee is T(t) = M - Ce-kt. Determine the value of C and K.
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