# Differential Eqns: Newton's Law of Cooling

ChemistryNat

## Homework Statement

A hot solid body is immersed into a large body of water
which is a 4 degrees C. After one hour the body's temperature is measured
at 40 degrees C, after 2 hours at 15 degrees. Assuming that Newton's law of
cooling applies and that the water body is mixed constantly and its temperature stays at 4 degrees, what was the temperature of the body initially?

## Homework Equations

Newton's Law of Cooling
dT/dt=κ(T-Tm)

## The Attempt at a Solution

Tm=4°C
T(1h)=40°C
T(2h)=15°C
T(0h)=?

dT/dt=k(T-Tm)
dT/dt=k(T-4)
separating variables and integrating yields
ln|T-4|= kt+c
T=ce^(kt)+4 .............eq1

T(1)=40 → 40=4+ce^k .....eq 2
T(2)=15 → 15=4+ce^2k ......eq3

So I solved for k using eq2 & eq3
36e^(2k)=11e^k
11/36=e^k
k=ln(11/36)

plugging k back into eq2
c=1296/11

eq 3 becomes :
T=(1296/11)e^(kt)+4

finding the initial temperature
T(0)=121.8°C

This answer just doesn't seem logical to me, a lot higher than I expected. I can't seem to figure out where I went wrong or if I did? Thank you! :)

iRaid
I'm not sure on the math, but your answer seems logical... If the water started at 4 degrees C and after an hour got to 40, the starting temperature had to be really high (since the temperature of the solid body is decreasing too).

But I'm not sure.

Staff Emeritus
Gold Member
2021 Award
I double checked for you and the final function that you got definitely matches the two data points they gave you, I think this is all correct.

If anything that number sounds way lower than I would have guessed actually, so that goes to show you how good intuition is at this kind of thing.

Homework Helper

Newton's Law of Cooling can be written in the form

T(t)-Tm=(T(0)-Tm)e-kt.

T(t) is the temperature of the body at time t, T(0) is the initial temperature, Tm is the temperature of the surroundings. You are given T(1) and T(2) and need to find T(0).

T(1)-Tm=(T(0)-Tm)e-k
T(2)-Tm=(T(0)-Tm)e-2k

square the first equation and divide by the second one.

ehild