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Newtons Law of Cooling.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \begin{document} \noindent Ex 2. Under appropriate laboratory conditions the temperature $T^\circ C$ of a beaker of a chemical\\ solution and the temperature $S^\circ C$ of a surrounding vat of cooler water satisfy\\ according to Newton's Law of cooling:\\ \begin{align} &\frac{dT}{dt}=-k(T-S)\qquad\text{where }k>0\:.\\ &\frac{dS}{dt}=0.75k(T-S) \end{align} \begin{align*} \text{(a)}\quad&\text{Show that $\frac{\:3\:}{4}\cdot\frac{dT}{dt}+\frac{dS}{dt}=0$ and hence deduce the result for $\frac{\:3\:}{4}T+S$.}\\ \text{Solution:}\quad&LHS=\frac{\:3\:}{4}\cdot\frac{dT}{dt}+\frac{dS}{dt}=-\frac{\:3\:}{4}k(T-S)+0.75k(T-S)=0\\ &\int_0^t\left(\frac{\:3\:}{4}\frac{dT}{dt}+\frac{dS}{dt}\right)\:dt=0\\ &\left[\frac{\:3\:}{4}T+S\right]_0^t =\left[\frac{\:3\:}{4}T+S\right]-\left[\frac{\:3\:}{4}T_0+S_0\right]=0\\ &\frac{\:3\:}{4}T+S=\frac{\:3\:}{4}T_0+S_0\\ \\ \text{(b)}\quad&\text{Find an expression for $\frac{dT}{dt}$ in terms of $T$, and}\\ &\text{hence show that $T=\frac{\:4\:}{7}C+Ae^{-\frac{7}{4}kt}$\:, where $C$ and $A$ are constants,}\\ &\text{satisfies this differential equation for any constant $A$}.\\ \text{Solution:}\quad&\text{From (a) }\frac{dS}{dt}=-\frac{\:3\:}{4}\frac{dT}{dt}\\ &\frac{d^2T}{dt^2}=\frac{d}{dt}\left(\frac{dT}{dt}\right) =\frac{d}{dt}\left[-k(T-S)\right] =-k\left(\frac{dT}{dt}-\frac{dS}{dt}\right) =-k\left(\frac{dT}{dt}+\frac{\:3\:}{4}\frac{dT}{dt}\right) =-\frac{\:7\:}{4}k\cdot\frac{dT}{dt}\\ &\frac{dT}{dt}=\int\left(-\frac{\:7\:}{4}k\cdot\frac{dT}{dt}\right)\:dt =-k\left(\frac{\:7\:}{4}T-C\right)=-\frac{\:7\:}{4}k\left(T-\frac{\:4\:}{7}C\right)\\ \therefore\quad&\frac{dT}{dt}=-\frac{\:7\:}{4}k\left(T-\frac{\:4\:}{7}C\right)\\ &\text{Given }T=\frac{\:4\:}{7}C+A\:e^{-\frac{7}{4}kt}\:,\\ &\frac{dT}{dt}=A\frac{d}{dt}e^{-\frac{7}{4}kt}=-\frac{\:7\:}{4}kA\:e^{-\frac{7}{4}kt} =-\frac{\:7\:}{4}k\left(T-\frac{\:4\:}{7}C\right)\:,\quad\text{which satisfies the differential equation.}\\ \\ \text{(c)}\quad&\ldots\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \end{align*} \noindent Ex 1. A cup of coffee cools at a rate proportional to the difference between its temperatures $T_c$ and that of its surroundings.\\ In winter, the room temperature is $15^\circ C$ and I must wait 10 minutes for my coffee to cool from $90^\circ C$ to $50^\circ C$.\\ \begin{align*} \text{(a)}\quad&\text{Explain why: }\frac{dT_c}{dt}=-k(T_c-15)\:.\\ \text{Solution:}\quad&\text{Newton's Law of Cooling \ldots duh!}\qquad(k>0)\\ \\ \text{(b)}\quad&\text{Show that: }T_c=15+75\left(\frac{7}{15}\right)^{\frac{t}{10}}\\ \text{Solution:}\quad&\text{Based on the ``duh'' rule: }T_c=15+A\:e^{-kt}\\ &\text{Given }A=90-15=75\text{ and when }t=10\:,\\ &T_c=15+75\:e^{-10k}=50\:,\quad e^{-10k}=\frac{35}{75}=\frac{7}{15}\:,\\ &e^{-kt}=\left(e^{-10k}\right)^\frac{t}{10}=\left(\frac{7}{15}\right)^\frac{t}{10}\:,\\ \therefore\quad&T_c=15+A\:e^{-kt}=15+75\:\left(\frac{7}{15}\right)^\frac{t}{10}\\ \\ \text{(c)}\quad&\text{How long must I wait, in summer, when the room temperature will be $25^\circ C$ to cool to $50^\circ C$?}\\ \text{Solution:}\quad&\text{So all the 15 is now 25, and $A=90-25=65$ .}\\ &T_c=25+65\:e^{-kt}=50\:,\quad e^{-kt}=\frac{25}{65}=\frac{5}{13}\:,\quad\left(\frac{7}{15}\right)^\frac{t}{10}=\frac{5}{13}\\ &t=10\cdot\frac{\ln\frac{5}{13}}{\ln\frac{7}{15}}=12.54\text{ minutes}=12:32\text{ minutes}\\ \end{align*} \end{document}