# 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

#### Solution

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:

### Formula

u(t) = T - (U - T)e^{(-(k)(t))}

Note: e represents Euler's number which is 2.71827

Solution

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}

2.71827

^{-3.45}

=.0317

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