If a cup of coffee has temperature 95∘C in a room where the temperature is 20∘C, then, according to Newton's Law of Cooling, the temperature of the coffee after t minutes is T(x)=20+75e^(-t/50). ?

Question

What is the average temperature (in degrees Celsius) of the coffee during the first half hour?

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Answer 1 · 2016-03-05T20:09:42

#76.4^@C#

We first find the temperature at time #t=0# to obtain

#T(0)=20+75e^(-0/50)=95^@C#.

Now after a half hour of cooling, we arrive at the temperature at #t=30# mins

#T(30)=20+75e^(-30/50)=61.161^@C#

Hence, he average temperature during the first half hour will be

#T_(avg)=1/(30-0)int_0^30(20+75e^(-t/50))dt#

#=1/30[20t-3750e^(-t/50)]_0^30#

#=1/30[600-2058.04+3750]#

#=76.4^@C#