The rate at which a body cools also depends on its exposed surface area S. If S is a constant, then a modification of (1) is dT kS(T – Tm) dt where k >0 and Tm is a constant. Suppose that two cups A and B are filled with coffee at the same time. Initially, the temperature of the coffee is 120° F. The exposed surface area of the coffee in cup B is twice the surface area of the coffee in cup A. After 30 min the temperature of the coffee in cup A is 90° F. If Tm =70° F, then what is the temperature of the coffee in cup B after 30 min?

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Q1. The mathematical formulation of Newton's empirical law of cooling/warming of an object is
given by the linear first-order differential equation
dT
= k(T – Tm)
(1)
dt
where k is a constant of proportionality, T(t) is the temperature of the object for t> 0, and Tm is
the ambient temperature that is, the temperature of the medium around the object.
The rate at which a body cools also depends on its exposed surface area S. If S is a constant, then
a modification of (1) is
dT
kS(T – Tm)
(2)
dt
where k> 0 and Tm is a constant. Suppose that two cups A and B are filled with coffee at the
same time. Initially, the temperature of the coffee is 120° F. The exposed surface area of the
coffee in cup B is twice the surface area of the coffee in cup A. After 30 min the temperature of
the coffee in cup A is 90° F. If Tm =70° F, then what is the temperature of the coffee in cup B
after 30 min?
Transcribed Image Text:Q1. The mathematical formulation of Newton's empirical law of cooling/warming of an object is given by the linear first-order differential equation dT = k(T – Tm) (1) dt where k is a constant of proportionality, T(t) is the temperature of the object for t> 0, and Tm is the ambient temperature that is, the temperature of the medium around the object. The rate at which a body cools also depends on its exposed surface area S. If S is a constant, then a modification of (1) is dT kS(T – Tm) (2) dt where k> 0 and Tm is a constant. Suppose that two cups A and B are filled with coffee at the same time. Initially, the temperature of the coffee is 120° F. The exposed surface area of the coffee in cup B is twice the surface area of the coffee in cup A. After 30 min the temperature of the coffee in cup A is 90° F. If Tm =70° F, then what is the temperature of the coffee in cup B after 30 min?
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