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.

 

[Office_Shredder]

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.

 

[ehild]

Your result is correct.

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

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

T(t) is the temperature of the body at time t, T(0) is the initial temperature, T_m 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