Early
in the morning the heater of a house breaks down. The outside
temperature is
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Solution: Click For PDF Version
According to
the Newton’s law of cooling if
where A is a constant. On integrating the above differential equation, we express its solution as
at a time t
later and
In our
problem the temperature of the surroundings, the outside
temperature,
The
temperature of the object, that is that of inside of the room, at
It is given
that the inside temperature drops to
We use this data for calculating the constant A.
We
will next find the additional time t for the inside
temperature of the room 18 to
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As a result, the inside temperature drops from
22 to
in 45 min. We
have to find how much longer will it take for the inside temperature
to fall by another
.
We can assume that the outside temperature
does not change and that the Newton’s law of cooling applies.
the difference in temperature between that of the object and the
surroundings (
)
is not too great, the rate of cooling of the object is proportional
to the temperature difference; that is,
is the temperature difference between the object and the
surroundings at
.
.
.
Therefore,
.
From the Newton’s law of cooling, we have