When an object is removed from a furnace and placed in an environment
with a constant temperature of 80F, its core temperature is 1500F. One
hour after it is removed, the core temperature is 1120F. What is the
core temperature 5 hours after the object is removed from the furnace?

Hi!!


According to the Newton's Law of Cooling we have that:

T(t) = Ts + (To - Ts)*e^(-k*t) ;

where:
t is the time in the preferred units (seconds, minutes, hours, etc.)
T(t) is the temperature of the object at time t
Ts is the sorrounding constant temperature
To is the initial temperature of the object
k is a constant to be found


According to the given data:
T(t) = 80 + (1500-80) * e^(-k*t) = 80 + 1420*e^(-k*t) 

We know that 1 hour after it is removed, the core temperature is 1120F, then:
T(1) = 80 + 1420 * e^(-k) = 1120

==> e^(-k) = (1120-80)/1420 = 0.7324
==> k = -ln(0.7324) = 0.3114

Now we have completed the object's cooling equation:
T(t) = 80 + 1420 * e^(-0.3114*t)

So we are able to find the object's temperature at time t=5 hours:
T(5) = 80 + 1420 * e^(-0.3114*5) =
     = 80 + 1420 * e^(-1.557) =
     = 80 + 1420 * 0.21 =
     = 378.2°F


For references about the derivation of the Cooling equation according
the Newton's Law of Cooling see:
"Newton's Law of Cooling":
http://www.sosmath.com/diffeq/first/application/newton/newton.html

"Newton's Law of Cooling -- from Eric Weisstein's World of Physics":
http://scienceworld.wolfram.com/physics/NewtonsLawofCooling.html


Search strategy:
"Newton's Law of Cooling"


I hope this helps you in the understanding of this topic, if you find
something unclear and/or incomplete please do not hesitate to request
for a clarification before rate this answer.


Regards,
livioflores-ga

Well explained answer and really helps me with my final exam study
guide.....that stressful time is rolling around again

I see that the new school semester has commenced.

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