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## 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! :)