{VERSION 3 0 "IBM INTEL NT" "3.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 256 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" 18 258 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" 18 260 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE " " -1 261 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" 18 263 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 265 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" 18 267 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" 18 269 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE " " -1 270 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 271 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 274 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 276 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 277 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE " " -1 279 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 280 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 281 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 282 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 283 "" 1 18 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 284 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 285 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 286 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 287 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE " " -1 288 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 289 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 290 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 291 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 292 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 293 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 294 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" 18 295 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 296 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE " " -1 297 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 298 "" 1 18 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 299 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 300 "" 1 18 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 301 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 302 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 303 "" 1 18 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 304 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 305 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE " " -1 306 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 307 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 308 "" 1 18 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 309 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 310 "" 1 18 0 0 0 0 0 0 1 0 0 0 0 0 0 }{CSTYLE "" -1 311 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 312 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 313 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 314 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE " " -1 315 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 316 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 " " 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {PSTYLE "Title" 0 18 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 1 0 0 0 0 0 0 }3 0 0 -1 12 12 0 0 0 0 0 0 19 0 }{PSTYLE "Author" 0 19 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 8 8 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 34 "Newton's Law Of Cooling-- Project 1" }}{PARA 19 "" 0 "" {TEXT -1 51 "by Dr. Fadyn--Southern Poly technic State University" }}{PARA 0 "" 0 "" {TEXT -1 23 " \+ " }{TEXT 309 48 " \+ " }{TEXT 310 15 " Cooling Coffee" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 244 "Newton's law of cool ing states that the rate of change in the temperature T(t) of a body i s proportional to the difference between the temperature of the surrou nding medium (the ambient temperature) and the temperature of the body . In symbols:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 " " }{XPPEDIT 258 0 "diff(T(t),t) = k*(A(t)-T(t));" "6#/-%%diffG6$-%\"TG6#%\"tGF**&% \"kG\"\"\",&-%\"AG6#F*F--F(6#F*!\"\"F-" }{TEXT -1 7 " " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 259 131 "In this project we make the assumption that the ambient temperatu re remains constant, so that A(t) = A and so our equation becomes:" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 " \+ " }{XPPEDIT 260 0 "diff(T(t),t) = k*( A-T(t));" "6#/-%%diffG6$-%\"TG6#%\"tGF**&%\"kG\"\"\",&%\"AGF--F(6#F*! \"\"F-" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 261 435 "You should know that this law works j ust as well for a heating object. If the ambient temperature is less \+ than the initial temperature of the object then the object cools and k >0. On the other hand, if the ambient temperature is greater than the initial temperature of the object, then the object heats and k<0. In any case, the differential equation is both separable and linear. Le t's see how to solve it as a separable equation:" }}{PARA 0 "" 0 "" {TEXT -1 27 " " }{TEXT 257 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 182 "dT/(A-T)=k*dt; #This command defines the different ial equation. Note that this explanation is a comment which you can i nclude in Maple by preceeding the comment with the symbol \"#\"." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 110 "eq:=Int(1/(A-t),T)=Int(k,t) +C; #Note that \"Int\" (with a capital I) will show, but not evaluate, the integrals." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "eq:=int(1/(A-T),T)=int(k,t)+C; #Not e that \"int\" (with a small i) actually evaluates the integrals." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 111 "exp(lhs(eq))=exp(rhs(eq)); \+ #(lhs and rhs refer to the \"left hand side\" and the \"right hand sid e\" respectively)." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 262 14 "If we replace " }{XPPEDIT 263 0 " exp(C);" "6#-%$expG6#%\"CG" }{TEXT 264 1 "\000" }{TEXT -1 0 "" }{TEXT 265 18 " with C we obtain:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "1/(A-T) = C*e^(k*t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "solve(%,T); # The % symbol in Maple refers to the previously computed result." }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 12 "simplify(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 266 35 "We will write this f inal result as:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 " " }{XPPEDIT 267 0 "T(t) = A+C* exp(-k*t);" "6#/-%\"TG6#%\"tG,&%\"AG\"\"\"*&%\"CGF*-%$expG6#,$*&%\"kGF *F'F*!\"\"F*F*" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 268 73 "If the initial condition \+ T(0) = T0 is given, then the solution becomes:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 " \+ " }{XPPEDIT 269 0 "T(t) = A+(T0-A)*exp(-k*t);" "6#/-%\"TG6#%\"tG,&% \"AG\"\"\"*&,&%#T0GF*F)!\"\"F*-%$expG6#,$*&%\"kGF*F'F*F.F*F*" }{TEXT -1 2 " " }{TEXT 270 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }{TEXT 271 59 "Of course Maple can easily provide the solution as follows:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 105 "de:=diff(T(t),t)=k*(A - T(t)); #Th is command assigns the variable name \"de\" to the differential equati on." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "ic:=T(0)=T0; #This c ommand assigns the variable name \"ic\" to the initial condition." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 118 "Temp:=dsolve(\{de,ic\},T(t) ); #Note the syntax of how Maple can be used to solve this first-order initial-value problem." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 272 358 "As a simple example, let 's consider a cup of coffee in an ordinary mug which is initially at a temperature of 180 degrees and which is set out in a room which maint ains a constant temperature of 70 degrees. Suppose that after 5 minut es the temperature of the coffee is 155 degrees. Let's set up the cor responding differential equation and try to find T(t)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "ode:=d iff(T(t),t)=k*(70-T(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "init:=T(0)=180;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "sol:=ds olve(\{ode,init\},T(t));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 274 94 "To find k we will use the fact that after 5 minutes the temperature is 155 degrees as follows: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 273 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "Coffeetemp:=t->70+110*exp(-k*t); #This comman d defines a function of t called \"Coffeetemp\"" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 22 "eq:=Coffeetemp(5)=155;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 15 "k:=solve(eq,k);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 114 "evalf(k); #This command is used to evaluate k to flo ating point form (that is to get a decimal approximation of k." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 103 "Coffeetemp(t); #Note that M aple substitutes the appropriate value of k into the function Coffeete mp(t)." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 275 149 "Now Coffeetemp is a function of t which \+ will give us the temperature of the coffee at any time t. For example , the temperature after ten minutes is:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "evalf(Coffeetemp(10)); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 276 0 "" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }{TEXT 277 35 "Let's look at a plot of Coffeetemp :" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 193 "plot(Coffeetemp(t),t=0..80); #Note the Maple syntax \+ for doing the plot of a function of one variable. A range of values fo r the independent variable (here t) must be given (here it is t=0..80) ." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 278 153 "As we would expect the temperature of the coffee \+ is approaching that of the room which is 70 degrees. In fact after ab out 90 minutes the temperature is:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "evalf(Coffeetemp(90));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 315 133 "As a fin al example, let's use Maple to help us find the value of t when the te mperature of the coffee is equal to 90 degrees. We do:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "solve( Coffeetemp(t)=90,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "eval f(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 316 557 "Observe that the \"solve\" command gives an exact answer (in terms of \"ln\") to the problem. We approximated this exact value with the command \"evalf\". In more difficult equat ions, sometimes Maple cannot obtain an exact solution to an equation. \+ In such cases, the \"fsolve\" command can sometimes give us an approx imate answer. The \"fsolve\" command requires a range of values in wh ich Maple searches for a solution. For example if we look at the plot above we can estimate that Coffeetemp(t) is equal to 90 when t is (ve ry roughly) 35 minutes, so we can do:" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "fsolve(Coffeetemp(t)=90 ,t=30..40);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 279 13 "Now for some " }{TEXT 280 8 "Question" }{TEXT 281 10 "s for yo u:" }{TEXT 285 0 "" }{TEXT 286 0 "" }{TEXT 287 0 "" }{TEXT -1 6 " \+ " }{TEXT 282 184 "Suppose the coffee in question (at 180 degrees init ially set out in a room at 70 degrees) is in an insulated mug so that \+ the temperature of the coffee after 5 minutes is 175 degrees. " }} {PARA 0 "" 0 "" {TEXT 293 1 " " }}{PARA 0 "" 0 "" {TEXT 288 70 "1. Fin d the function which will predict the temperature at any time. " }} {PARA 0 "" 0 "" {TEXT 312 1 " " }}{PARA 0 "" 0 "" {TEXT 289 45 "2. Fin d the temperature after half an hour. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 290 60 "3. Graph the temperature function o ver the first two hours. " }}{PARA 0 "" 0 "" {TEXT 313 2 " " }}{PARA 0 "" 0 "" {TEXT 291 9 "4. Graph " }{TEXT 283 4 "both" }{TEXT 284 76 " \+ temperature functions on the same coordinate axes for comparison purpo ses. " }}{PARA 0 "" 0 "" {TEXT 314 1 " " }}{PARA 0 "" 0 "" {TEXT 292 222 "5. Find the difference between the two coffee temperatures after 1 hour. Is it true that the temperature of the coffee in the ordinar y mug is always (for t>0) less than the temperature of the coffee in t he insulated mug?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 294 138 "6. Compare the values of k for each \+ situation. Looking at these and keeping in mind the general solution \+ for Newton's law of cooling:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 " " }{XPPEDIT 295 0 "T( t) = A+exp(-k*t)*(-A+T0);" "6#/-%\"TG6#%\"tG,&%\"AG\"\"\"*&-%$expG6#,$ *&%\"kGF*F'F*!\"\"F*,&F)F2%#T0GF*F*F*" }{TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 296 7 "make a \+ " }{TEXT 298 7 "general" }{TEXT 297 83 " statement concerning the size of k related to the rate at which an object cools." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 299 14 "7. Gr aph the " }{TEXT 300 11 "derivatives" }{TEXT 301 154 " of both tempera ture functions on the same coordinate axes. Find the time(s) when bot h derivatives are equal accurate to at least three decimal places. " }{TEXT 303 4 "Hint" }{TEXT 304 174 ": You can use Maple's \"fsolve\" \+ command to accomplish this. Look up the syntax in Maple \"Help\". Is \+ it true that the coffee in the standard mug is always cooling at a fas ter " }{TEXT 308 4 "rate" }{TEXT 307 38 " than the coffee in the insul ated mug?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 302 180 "8. Based on your answers to Question 7, do you f ind it necessary to amend the general statement you made in Question 6 in any way? If so, how? Make a new statement if necessary." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 305 371 "9. Find a value of k so that the temperature of the coffee after 5 minutes is 165 degrees. Your answer should be accurate to at least five decimal places. You should assume, of course, that the co ffee temperature is originally 180 degrees and that it is set out in a room which maintains a constant temperature of 70 degrees. Let's cal l this new mug semi-insulated." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 306 504 "10. Assume that coffee \+ is originally set out in the ordinary mug, and after ten minutes it is transferred to the semi-insulated mug, and after another ten minutes \+ it is transferred again into the insulated mug. Let's say in addition that each time the coffee is transferred it suffers a 3 degree heat l oss. Find the temperature of the coffee after a half hour (that is a \+ half hour after it is originally set out in the ordinary mug) to the n earest hundredth of a degree if no more transfers are made." }}{PARA 0 "" 0 "" {TEXT -1 105 "______________________________________________ ___________________________________________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 311 44 " " }}}}{MARK "0 0 0" 0 } {VIEWOPTS 1 1 0 1 1 1803 }