Question

According to Newton's Law of Cooling

A cup of coffee with temperature of 130F is placed in a freezer with temperature 0F. After 5 minutes, the temperature of the coffee is 87F. Find the coffee's temperature

after 10 minutes.

Answer #1

Answer:- T(10) = 58.237

According to Newton's Cooling Law

T(t) =
T_{a}+(T_{o}-T_{a})e^{-kt}

where:

t is Time

T_{a} is the (constant) ambient temperature

T_{o} is the initial temperature of the object

k is a proportionality constant

Given:-

T(5 Minutes) = 87

T_{a} = 0F

T_{o} = 130F

T(5) = 87 = 0 + (130 - 0 )e^{-k(5)}

I will solve this for k (in the units I have chosen):

87 = 0 + 130e^{-5k}

0.6692 = e^{-5k}

ln(0.6692) = lne^{-5k}

-0.40167 = -5k

k = 0.08033

Then at t = 10 minutes

T(t) = 130e^{-(0.08033)t}

T(10) = 130e^{-(0.08033)10}

^{T(10) = 58.237F}

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