Newton's Law of Cooling as a Differential Equation dT 4 = k(M – T) dt !! where T'is the temperature of the object at a given time t, M is the temperature of the surrounding medium, and k is a positive constant. From this we note that if M>T, we have heating, since M – T > 0 thus dT/dt > 0, which means an increasing T, heating up! If M
Newton's Law of Cooling as a Differential Equation dT 4 = k(M – T) dt !! where T'is the temperature of the object at a given time t, M is the temperature of the surrounding medium, and k is a positive constant. From this we note that if M>T, we have heating, since M – T > 0 thus dT/dt > 0, which means an increasing T, heating up! If M
Elements Of Electromagnetics
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![Newton's Law of Cooling as a Differential Equation
dT
k(м-т)
dt
where Tis the temperature of the object at a given time t, M is the temperature of
the surrounding medium, and k is a positive constant. From this we note that if
M>T, we have heating, since M – T > 0 thus dT/dt > 0, which means an increasing T,
heating up! If M<T, we have cooling, as since M – T < 0 thus dT/dt < 0, which means
a decreasing T, cooling down!
Now, working from the differential equation of Newton's law of cooling, solve the
following situation:
Mary and Bob each bring a cup of coffee from the dining hall at Christchurch to their
class, the classroom that is kept at a constant temperature of 70 °F. Assume the
temperature of the coffee at the beginning of class is 200 °F. Mary has a well-
insulated coffee mug; her coffee is 194 °F after 5 minutes of class. Bob has a paper
cup coffee mug, it is not well insulated; his coffee is 150 °F after 5 minutes.
Calculate the temperature of Mary's coffee at the end of class.
Calculate the temperature of Bob's coffee at the end of class.
Comment on the value of the constant usually referred to as k in the differential
equation, how does this value relate to the heat transfer between the coffee and
surrounding medium?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7269e1e1-4eee-4be5-b1eb-96a63ee67eff%2F80366d80-7973-4096-8bf0-cba4d4583a74%2Fbcr98uq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Newton's Law of Cooling as a Differential Equation
dT
k(м-т)
dt
where Tis the temperature of the object at a given time t, M is the temperature of
the surrounding medium, and k is a positive constant. From this we note that if
M>T, we have heating, since M – T > 0 thus dT/dt > 0, which means an increasing T,
heating up! If M<T, we have cooling, as since M – T < 0 thus dT/dt < 0, which means
a decreasing T, cooling down!
Now, working from the differential equation of Newton's law of cooling, solve the
following situation:
Mary and Bob each bring a cup of coffee from the dining hall at Christchurch to their
class, the classroom that is kept at a constant temperature of 70 °F. Assume the
temperature of the coffee at the beginning of class is 200 °F. Mary has a well-
insulated coffee mug; her coffee is 194 °F after 5 minutes of class. Bob has a paper
cup coffee mug, it is not well insulated; his coffee is 150 °F after 5 minutes.
Calculate the temperature of Mary's coffee at the end of class.
Calculate the temperature of Bob's coffee at the end of class.
Comment on the value of the constant usually referred to as k in the differential
equation, how does this value relate to the heat transfer between the coffee and
surrounding medium?
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