Question
Consider a 92∘C cup of coffee placed in a 24∘C room. Suppose it is known that the coffee cools at a rate of 2∘C/min when it is 70∘C. Then according to Newton's law of cooling, the temperature T(t) of the coffee t minutes after being placed in the room satisfies the differential equation,
dT/dt=(2/(70−24))(24−T).
What is the temperature (in Celcius) of the coffee at time t=1 min?
I tried a few methods but I am completely stumped with this questions. Can someone please show me steps on how to do it
dT/dt=(2/(70−24))(24−T).
What is the temperature (in Celcius) of the coffee at time t=1 min?
I tried a few methods but I am completely stumped with this questions. Can someone please show me steps on how to do it
Answers
They gave you the hint "Newton's Law of Cooling" so you do not even have to do the calculus.
for example:
http://www.softschools.com/formulas/physics/newtons_law_of_cooling_formula/93/
for example:
http://www.softschools.com/formulas/physics/newtons_law_of_cooling_formula/93/
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