Sammy's question at Yahoo Answers regarding Newton's Law of Cooling

[MarkFL]

Here is the question:

First order differential equation problem...?

rate of change of temperature between an object and its surrounding medium is proportional to the temperature difference. A hot coal of 160 degree celsius is immersed in water of 5 degree celsius. After 40 seconds, the temperature of coal is 70 degree celsius. Assume water is kept at constant temperature. Find
(a) the temperature of coal after 1 minute
(b) When does the temperture of coal reach 20 degree celsius

I have posted a link there to this topic so the OP can find my work.

 

[MarkFL]

Re: sammy's question at Yahoo Answers regarding Newton's Law of Cooling

Hello sammy,

Newton's law of Cooling states that the time rate of change of the temperature $T$ of an object is proportional to the difference between the ambient temperature $M$ and the temperature of the object. Stated mathematically, this is:

\(\displaystyle \frac{dT}{dt}=-k(T-M)\) where \(\displaystyle T(0)=T_0,\,0<k\in\mathbb{R}\) and \(\displaystyle T>M\).

The ODE is separable and may be written:

\(\displaystyle \frac{1}{T-M}\,dT=-k\,dt\)

Integrating, using the boundaries, and dummy variables of integration, we find:

\(\displaystyle \int_{T_0}^{T(t)}\frac{1}{u-M}\,du=-k\int_0^t v\,dv\)

(1) \(\displaystyle \ln\left(\frac{T(t)-M}{T_0-M} \right)=-kt\)

If we know another point \(\displaystyle \left(t_1,T_1 \right)\) we may now find $k$:

\(\displaystyle -k=\frac{1}{t_1}\ln\left(\frac{T_1-M}{T_0-M} \right)\)

Hence, we find:

(2) \(\displaystyle T(t)=\left(T_0-M \right)\left(\frac{T_1-M}{T_0-M} \right)^{\frac{t}{t_1}}+M\)

(3) \(\displaystyle t=\frac{t_1\ln\left(\frac{T(t)-M}{T_0-M} \right)}{\ln\left(\frac{T_1-M}{T_0-M} \right)}\)

We may now use the given data to answer the questions:

\(\displaystyle T_0=160,M=5,t_1=40,T_1=70\)

(a) Using (2) and $t=60$, we find:

\(\displaystyle T(60)=\left(160-5 \right)\left(\frac{70-5}{160-5} \right)^{\frac{60}{40}}+5=\left(155 \right)\left(\frac{13}{31} \right)^{\frac{3}{2}}+5=5\left(1+13\sqrt{\frac{13}{31}} \right)\approx47.09244817717765\)

(b) Using (3) and $T(t)=20$, we find:

\(\displaystyle t=\frac{40\ln\left(\frac{20-5}{160-5} \right)}{\ln\left(\frac{70-5}{160-5} \right)}=\frac{40\ln\left(\frac{3}{31} \right)}{\ln\left(\frac{13}{31} \right)}\approx107.49243770293884\)