A hot drink is taken outside on a cold winter day when the air temperature is –6°C. According to a principle of physics called Newton's Law of Cooling, the temperature T (in degrees Celsius) of the drink t minutes after being taken outside is given by T(t) = -6 + Ae-kt. %3D where A and k are constants. (a) Suppose that the temperature of the drink is 86°C when it is taken outside. Find the value of the constant A. (b) In addition, suppose that 20 minutes later the drink is 30°C. Find the value of the constant k. (c) What will the temperature be after 32 minutes?

University Physics Volume 2
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Chapter2: The Kinetic Theory Of Gases
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A hot drink is taken outside on a cold winter day when the air temperature is –6°C. According to a principle of
physics called Newton's Law of Cooling, the temperature T (in degrees Celsius) of the drink t minutes after
being taken outside is given by
T(t) = -6 + Ae-kt.
%3D
where A and k are constants.
(a) Suppose that the temperature of the drink is 86°C when it is taken outside. Find the value of the constant A.
(b) In addition, suppose that 20 minutes later the drink is 30°C. Find the value of the constant k.
(c) What will the temperature be after 32 minutes?
Transcribed Image Text:A hot drink is taken outside on a cold winter day when the air temperature is –6°C. According to a principle of physics called Newton's Law of Cooling, the temperature T (in degrees Celsius) of the drink t minutes after being taken outside is given by T(t) = -6 + Ae-kt. %3D where A and k are constants. (a) Suppose that the temperature of the drink is 86°C when it is taken outside. Find the value of the constant A. (b) In addition, suppose that 20 minutes later the drink is 30°C. Find the value of the constant k. (c) What will the temperature be after 32 minutes?
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