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chapter2.3p
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à 2.3èNewën's Law ç Coolïg
äè Solve ê problem.
âè A hot object at 100⌡F is placed ï a 60⌡F room. If its temp-
erature is 90⌡F after 10 mïutes,èwhat will its temperature
be ï an hour?èWith T╠ = 100⌡F å T╬ = 60⌡F, ê temperature
functionèT = (T╠-T╬)eúÖ▐ + T╬ becomes T = 40eúÖ▐ + 60.è
Subsitutïg T = 90⌡F at t = 10 mï å solvïg yields
s = 0.0287 mïúî so T = 40eúò°òìôÆ▐ + 60.èSubstitutïg
t = 60 mï yields T = 40eúò°òìôÆÑæòª + 60 = 67.1⌡F
éSèèWhen a hot object is placed ï an environment that has
a constant, cooler temperature, ê object's temperature falls
as described by NEWTON'S LAW OF COOLING
dT
────è=è- s(T - T╬)
dt
whereèt = ê time sïce ê object was placed ï ê room
èèè T = ê object's temperature at time t
èèè T╬ = ê room's ambient temperature
èèè s = proportionality constant
è This situation can be described by ê INITIAL VALUE problem
dT
────è=è- s(T - T╬)
dt
T(0)è=èT╠
è This is a SEPARABLE differential equation which yields ê
separated ïtegrals
░èèdTèèèè ░
▒è──────è =è ▒ -s dt
▓è T-T╬èèèè▓
This ïtegrates ë
ln[T-T╬]è=è-stè+èln[C]èC constant ç ïtegration
èíèT-T╬è┐
ln│ ────── ▒è=è-st
èèèèè└è Cèè┘
Exponentiatïg both sides
T-T╬
──────è=èeúÖ▐
è C
Or
T - T╬è=èCeúÖ▐
Tè=èCeúÖ▐ + T╬
To evaluate ê constant ç ïtegration C, substitute 0 for t
å T╠ for T ë yield
T╠è=èC + T╬
soèèèCè=èT╠ - T╬
Thus ê TEMPERATURE FUNCTION becomes
Tè=è(T╠-T╬)eúÖ▐ + T╬
èèAs t goes ë ïfïity, ê exponential goes ë zero
leavïg ê object at room temperature.
èèIn usïg ê temperature function ë fïd temperatures
at various times, values for ê ïitial temperature T╠ å
ê room temperature T╬ are usually given, but this still
leaves ê parameterèsèë be found before ê temperature
function can be used.èThis is normally done by measurïg ê
object's temperature at a later time.èThis ïformation is
substituted ïë ê temperature function which is ên
solved for s.
èèThe situation that a hot object is placed ï a cooler
room could be reversed ë brïgïg a cold object ïë a
warmer room.èThis would be described by "Newën's Law ç
Heatïg" which is still described by ê differential equation.
dT
────è=è- s(T - T╬)
dt
In this situtation ê quantity ï parenêses is now nega-
tive makïg ê right hå side positive i.e. ê object's
temperature will rise.èThe temperature function
Tè=è(T╠-T╬)eúÖ▐ + T╬
is still valid with ê note that ê quantity ï parenêses
is negative so T will go from its ïitial value ç T╠ ë its
fïal, higher temperature ç T╬
è1èèThe temperature ç a hot liquid is 100⌡F å it is
placed ï a 70⌡F room.èAfter 10 mïutes ê temperature is
90⌡F.èUse this ïformation ë fïd s ï Newën's Law ç
Coolïg temperature function.
è A) 4.05 mïúîèB)è0.405 mïúîèC)è0.0405 mïúîèD) 0.00405 mïúî
ü èèSubstitutïgèT╠ = 100⌡F, T╬ = 70⌡F , T = 90⌡F at t = 10
mïutes ïë ê temperature function gives
90è=è30eúîòÖ + 70
20è=è30eúîòÖ
orèèè2/3è=èeúîòÖ
Takïg ê natural log ç both sides yields
ln[2/3]è=èln[eúîòÖ]è=è-10s
Thusèèsè=è- ln[2/3] / 10è=è0.0405 mïúî
ÇèC
è2èèThe temperature ç a hot liquid is 100⌡F å it is
placed ï a 70⌡F room.èAfter 10 mïutes ê temperature is
90⌡F.èNewën's Law ç Coolïg temperature function is
T = 30eúò°òÅòÉ▐ + 70.èFïd ê temperature at 30 mïutes
after ê object has been ïtroduced ë ê room.
A)è 73.4⌡FèèB)è75.0⌡Fè C)è76.2⌡Fè D)è78.9⌡F
ü The temperature function is
T = 30eúò°òÅòÉ▐ + 70
Substitutïg t = 30 mïutes yields
T = 30eúò°òÅòÉÑÄòª + 70è=è78.9°F
Çè D
3èThe temperature ç a hot liquid is 100⌡F å it is
placed ï a 70⌡F room.èAfter 10 mïutes ê temperature is
90⌡F.èNewën's Law ç Coolïg temperature function is
T = 30eúò°òÅòÉ▐ + 70.èHow long after ê object has been
ïtroduced ë ê room will its temperature be 80⌡F?
A) 20 mïutesèB)è22 mïutesèC)è25 mïutesèD)è27 mïutes
üèThe temperature function is
T = 30eúò°òÅòÉ▐ + 70
Substitutïg t = 80⌡F yields
80 = 30eúò°òÅòÉ▐ + 70è
or
10 = 30eúò°òÅòÉ▐
1/3 = eúò°òÅòÉ▐
Takïg ê natural log ç both sides gives
ln[1/3]è=èln[eúò°òÅòÉ▐]è=è-0.0405t
Thus
tè=è- ln[1/3] / 0.0405è=è27 mïutes
Ç D
è4èA êrmometer is taken from ê outside on a warm day
(100⌡F) å placed ï a freezer.èAfter 15 mïutes, ê temp-
erature reads 80⌡F å after 30 mïutes it reads 65⌡F.èFïd
ê temperature ç ê freezer
A)è25⌡FèèèB)è20⌡FèèèC)è15⌡FèèèD)è10⌡F
üèSubstitutïg ê first readïg ïë ê temperature function
yields
80 = (100-T╬)eúîÉÖ + T╬
Substitutïg ê second readïg gives
65 = (100-T╬)eúÄòÖ + T╬
Asè eúÄòÖè= [ eúîÉÖ ]ì solvïg for those quantities will allow
a relation between ê two readïgs ë be found
eúÄòÖè=è(65 - T╬) / (100 - T╬)
å eúîÉÖè=è(80 - T╬) / (100 - T╬)
Thus, byèeúÄòÖè= [ eúîÉÖ ]ì
è65 - T╬èèèíèè80 - T╬è ┐ 2
──────────è=è▒è──────────è│
100 - T╬èèè└è 100 - T╬è ┘
Squarïg å cross multiplyïg gives
(65 - T╬)(100 - T╬)è=è(80 - T╬)ì
Squarïg
6500 - 165T╬ + T╬ìè=è6400 - 160T╬ + T╬ì
Simplifyïg
100è=è5T╬
Soèè T╬ = 20⌡F
ÇèBè
è5è A êrmometer is taken from a refrigeraër at 40⌡F å
taken outside on a 100⌡F day.èIf it takes 5 mïutes ë go
ë 60⌡F, how long will it take ë go ë 90⌡F?
A) 16.1 mïèèèB) 19.1 mïèè C)è22.1 mïèèèD)è25.1 mï
üèèèSubstitutïg ê basic ïformation that T╠ = 40⌡F,
T╬ = 100⌡ ïë ê temperature function yields
T = -60eúÖ▐ + 100
èèTo evaluate s, substitute T = 60⌡F at t = 5 mïutes ïë
ê temperature function.
60 = -60eúÉÖ + 100
orèèè-40 = -60eúÉÖ
2/3 = eúÉÖ
Takïg ê natural log ç both sides
ln[2/3] = ln[eúÉÖ] = -5s
soèèès = - ln[2/3] / 5 mï =è0.0811 mïúî
Thus ê temperature function becomes
T =è-60eúò°òôîî▐ + 100
Substitutïg T = 90⌡F yields
90è=è-60eúò°òôîî▐ + 100
-10è=è-60eúò°òôîî▐
1/6è=èeúò°òôîî▐
Takïg ê natural log ç both sides gives
ln[1/6]è=èln[eúò°òôîî▐]è=è-0.0811t
Thus
tè=è- ln[1/6] / 0.0811è=è22.1 mïutes
Ç C
è6èèA coroner arrives at a murder at 9 p.m. å fïds ê
temperature ç ê body is 85⌡F ï a room whose constant
temperature is 60⌡F.èAn hour later, ê body's temperature
is down ë 80⌡F.èWhen did ê murder occur?
A)è3:30 p.m.è B)è4:30 p.m.è C)è5:30 p.m.è D)è6:30 p.m.
üèè In this situtation, êre are 2 temperatures taken at
specified times.èIt is given that T╬ = 60⌡F.èThe ïitial
temperature ç ê body can be assumed as T╠ = 98.6⌡F unless
êre is evidence that ê victim had a fever.èWe need ë
fïd ê time ç ê murder, let
t = time from ê murder ë 9 p.m. so
t + 1 = time ç second temperature measurement
or 10:00 p.m.
èè Substitution gives ê temperature function at 9:00 p.m.
85 = 38.6eúÖ▐ + 60
orèèè25 =è38.6eúÖ▐
èè At 10:00 p.m., ê temperature funciën is
80 = 38.6eúÖÑ▐óîª + 60
or 20 = 38.6eúÖÑ▐óîª
Dividïg ê ëp equation by ê botëm yields
25èèèè 38.6eúÖ▐
────è=è──────────────
20èèè 38.6eúÖÑ▐óîª
Simplifyïg
1.25è=èeúÖ▐úÑúÖÑ▐óè=èeÖ
Takïg ê natural log ç both sides
ln[1.25]è=èln[eÖ] = s
Thusèèès =è0.0223 hourúî
Substitutïg ïë ê time function for 9 p.m. gives
85 =è38.6eúò°ììÄ▐ + 60
orèèè25è=è38.6eúò°ììÄ▐
25/38.6è=èeúò°ììÄ▐
Takïg ê natural log ç both sides yields
ln[25/38.6]è=èln[eúò°ììÄ▐]è=è-0.223t
Thusèè tè=è- ln[25/38.6] / 0.223è=è3.54 hours
è So 9 p.m. is 3.5 hours after ê murder, so ê
murder was committed at 5:30 p.m.
ÇèC