Newton's Law of Cooling Calculator
Newton's Law of Cooling Sample ProblemA cup of of tea begins with a temperature of 140.0 F, and the surrounding air is 80.0 F. If it cools at a constant k = 0.0023 1/s, what will the temperature of the cup of tea be after 25 minutes?
In this case we are solving for u at time t, therefore:
Formulau(t) = T - (U - T)e(-(k)(t))
Note: e represents Euler's number which is 2.71827
The tea cools for 25 minutes:
t = 25 × 60 seconds (per min)
t = 1500 seconds
u(1500 seconds) = 80F + (140F - 80F)e(-(.0023)(1500))
u(1500 seconds) = 80F + (140F - 80F)e(-3.45)
next lets solve for e-3.45
u(1500 seconds) = 80F + (140F - 80F)(.0317)
u(1500 seconds) = 80F + 1.9048
u(1500 seconds) = 81.905F
The cup of tea will be 81.9F at 25 min (or in 1500 seconds).
Below is a very good explanation of Newton's Law of Cooling