**BRAVO! That is
correct.**

A cup of hot tea, initially at 95^{o}C, cools to
75^{o}C, in 5.0 minutes when sitting in 25^{o}C
sirrpimdomgs. Use Newton's Law of Cooling to determine how long it
will take to cool to 35^{o}C.

Newton's Law of Cooling is

T_{surf} is T_{surf}(t), the temperature
of the surface as a function of time t.
T_{o} is the temperature of the surroundings,
T_{o} = 25^{o}C .

T
is the __initial__ temperature difference; T
= 95^{o}C - 25^{o}C = 70 C^{o}.

We know T_{surf} at time t = 5.0 min; that is
75^{o}C.

T_{surf} = T_{o} + T
e^{ }^{- t}^{/}
75^{o}C = 25^{o}C + (70^{o}C)e^{
}^{- 5 min}^{/}

We know everything in this equation except the time constant
.
This describes the cooling rate and is determined by the physical
situation. (Solving for
gives opportunity to play with the natural logrithm and exponential
functions on a calculator).

(70^{o}C)e^{ }^{- 5
min}^{/
}= 75^{o}C - 25^{o}C = 50^{o}C
e^{ }^{- 5
min}^{/
}= 50^{o}C/70^{o}C

e^{ }^{- 5
min}^{/
}= 0.714

How can we get
out of this exponent? Recall -- from a long-ago algebra class -- that
ln(e^{x}) = x. That means that

ln(e^{ }^{- 5
min}^{/
}) = - 5 min/
So we take the ln() -- the "natural logrithm" function -- of both
sides of the equation,

ln(0.714) = - 0.3368
- 5 min/
= - 0.3368

5 min/
= 0.3368

/5
min = 1/0.3368

/5
min = 2.967

= (2.967)(5 min )

= 14.84 min

This is the **time constant** for the exponent in Newton's Law
of Cooling. Now we can evaluate T_{surf} for __any__ time
t. Or we can calculate the time t for __any__ temperature
T_{surf}. What time t gives T_{surf} =
35^{o}C?

T_{surf} = T_{o} + T
e^{ }^{- t}^{/}
35^{o}C = 25^{o}C +
(70^{o}C)e^{
}^{-
t}^{/14.84 min}

35^{o}C - 25^{o}C =
(70^{o}C)e^{
}^{-
t}^{/14.84 min}

10^{o}C = (70^{o}C)e^{
}^{-
t}^{/14.84 min}

(70^{o}C)e^{
}^{-
t}^{/14.84 min }=
10^{o}C

e^{ }^{-
t}^{/14.84 min }=
10^{o}C/70^{o}C

e^{ }^{-
t}^{/14.84 min }=
0.143

And now -- just as before -- we need to use the ln() -- the
"natural logrithm" function -- to get the time t out of the
exponent,

ln(e^{ }^{-
t}^{/14.84 min})^{
}= ln( 0.143)
- t/14.84 min = - 1.95

- t/14.84 min = - 1.95

- t/14.84 min = - 1.95

t/14.84 min = 1.95

t = (1.95)(14.84 min)

**t **** ****= 28.9 min**

(c) 2000, Doug Davis; all rights reserved.