Newton's Law of Cooling Calculator

T is the constant temperature of the surrounding medium

U is the temperature of the heated object at t = 0

k is the constant cooling rate, enter as positive as the calculator considers the negative factor

t is the time that has elapsed since object u had it's temperature checked


u(t) is the temperature of object u at time t

Newton's Law of Cooling Sample Problem

A 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:


u(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

bizSinny.com - Our Calculator Micro Site

bizSkinny.com provides easy to use quick reference online calculators that solve basic or scientific problems for school, business or personal use. I have been working on this site for a long time now, and while there are still a lot of calculators that I haven't created yet we offer quite a few. So far we cover the following areas of math; algebra, geometry, finance, health, statistics, math, loans, and more. Our calculators are online for free! If you have a suggestion, please contact us.