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Example 2.13 Speed of cooling of body in mid air
proportional to the difference between the temperature of body

and temperature of air. The temperature of air is evened 20 С.
It is known, that during 20 minutes a body cools down from


100 to 60 . To set the law of change of temperature of body
depending on time.

 If to note time through t , and temperature of body
through U , speed of cooling of body, or in other words, speed
dU
of change of his temperature will be derivative . After
dt
problem specification we have
dU
 ( k U  20) ,
dt
where k – coefficient of proportion. Separating variables, we
have
dU
 kdt .
U  20
Integrating left and right parts, we will have

dU

  k dt  ln C
U  20

(as we will potencies in future, it is here comfortably to write
lnC in place of С )
or U  20 Ce kt .
For determination of constant C and k we will use
problem specifications :


0
U  100 at t  and U  60 at t  20 . First it is
accepted to name from these terms „initial", with its help we
find C . The second condition helps to set the coefficient of
proportion k :
 100  20 C
 20k
 60  20 Ce

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