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:
$$\frac{dT}{dt}=-k(T-M)$$ where $$T(0)=T_0,\,0<k\in\mathbb{R}$$ and $$T>M$$.
The ODE is separable and may be written:
$$\frac{1}{T-M}\,dT=-k\,dt$$
Integrating, using the boundaries, and dummy variables of integration, we find:
$$\int_{T_0}^{T(t)}\frac{1}{u-M}\,du=-k\int_0^t v\,dv$$
(1) $$\ln\left(\frac{T(t)-M}{T_0-M} \right)=-kt$$
If we know another point $$\left(t_1,T_1 \right)$$ we may now find $k$:
$$-k=\frac{1}{t_1}\ln\left(\frac{T_1-M}{T_0-M} \right)$$
Hence, we find:
(2) $$T(t)=\left(T_0-M \right)\left(\frac{T_1-M}{T_0-M} \right)^{\frac{t}{t_1}}+M$$
(3) $$t=\frac{t_1\ln\left(\frac{T(t)-M}{T_0-M} \right)}{\ln\left(\frac{T_1-M}{T_0-M} \right)}$$
We may now use the given data to answer the questions:
$$T_0=160,M=5,t_1=40,T_1=70$$
(a) Using (2) and $t=60$, we find:
$$T(60)=\left(160-5 \right)\left(\frac{70-5}{160-5} \right)^{\frac{60}{40}}+5=\left(155 \right)\left(\frac{13}{31} \right)^{\frac{3}{2}}+5=5\left(1+13\sqrt{\frac{13}{31}} \right)\approx47.09244817717765$$
(b) Using (3) and $T(t)=20$, we find:
$$t=\frac{40\ln\left(\frac{20-5}{160-5} \right)}{\ln\left(\frac{70-5}{160-5} \right)}=\frac{40\ln\left(\frac{3}{31} \right)}{\ln\left(\frac{13}{31} \right)}\approx107.49243770293884$$