Answer to Question #293572 in Differential Equations for kor

Question

A body at an unknown temperature is placed in a room which is held

at a constant temperature of 32° F. If after 10 minutes the temperature of the

body is 0° F and after 20 minutes the temperature of the body is 10° F, find the

unknown initial temperature.

Answer 1

Solution;

From Newton's Law of Cooling ,the Rate of cooling of the body is proportional to difference between the body's temperature and the surrounding temperature,given as;

$\frac{dT}{dt}=-k(T-T_s)$

where T is the body's temperature and k is the proportionality constant.

$T_s=32°F$

$\frac{dT}{dt}=-k(T-32)$

Seperate by variables;

$\frac{dT}{T-32}=-kdt$

Integrate;

$ln(T-32)=-kt+C$

Rewrite;

$T-T_s=e^{-kt+C}$

$T=T_s+Ce^{-kt}$

At t=0;

$C=T_0-T_s$

Where $T_0$ is the initial temperature.

The equation becomes;

$T=T_s+(T_0-T_s)e^{-kt}$

Given;

t=10,T=0°F

By substitution;

$0=32+(T_0-32)e^{-20k}$

$(32-T_0)e^{-10k}=32$

$32-T_0=\frac{32}{e^{-10k}}$

Also;

t=20,T=10°F

$10=32+(T_0-32)e^{-20k}$

$(32-T_0)e^{-20k}=22$

Rewrite;

$\frac{32}{e^{-10k}}e^{-20k}=22$

$32e^{-10k}=22$

$-10k=ln(\frac{22}{32})$

$k=-0.1ln(\frac{22}{32})$

Therefore;

$32-T_0=\frac{32}{e^{-10k}}$

$32-T_0=\frac{32}{0.6875}$

$T_0=-14.5°F$