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• Diego Marino

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UNIVERSIDAD NACIONAL DE TRUJILLO Facultad de Ingeniería Escuela de Ingeniería de Sistemas

Practica 3

ALUMNO: MARINO RAMOS DIEGO ALEJANDRO Curso: Matemática Aplicada Docente: Luis Lara Ciclo: III 2020

1. Cooling of a cake. When a cake is removed from an oven, its temperature is measured at 300oF. Three minutes later, its temperature is 200°F. How long will it take the cake to cool off to a room temperature of 70°F? T ( 0 )=300 T ( 3 )=200 dT =k ( T−70 ) , dt dT

∫ T −70 =∫ k dt ln |T −70|=kt +c T =70+c 2 e kt T ( 0 )=300=70+ c 2 c 2=230 T ( 3 )=200=70+ 230 c2 e 3 k e3 k=

13 23

K=−0.19018 T (t)=70+ 230 e−0.19018 t

T ( t )=70=70+230 e−0.19018t t ≈ 30’

2. Thermometer. A thermometer is removed from a room where the temperature is 70°F and is taken outside, where the air temperature is 10°F. After one-half minute, the thermometer reads 50°F. What is the reading of the thermometer at t = 1 min? How long will take the thermometer to reach 15°F?

dT dT dT =k ( T −T m )=¿ =kdt=¿∫ =∫ kdt dt (T −T m) (T −T m ) ln |T −70|=kt +c=¿T =T m +C ekt

Cuando temperatura del aire: 10°F entonces T m=10 ° F T(0) = 70°F T(0) = 70 = 10 + C

=>

C = 60°F

T =10+60 e kt Termometro marca 50°F luego de medio minuto 50=10+60 e 0.5 k 0.5 k =ln

( 6040 )=¿ k=−0.8109=¿ T =10+60 e

−0.8109t

Cuando el termometro alcanza 15°F: 15=10+60 e−0.8109t

e−0.8109 t=

5 =¿ t ≈ 3.06 ' 60

3. Thermometer. A thermometer is taken from an inside room to the outside, where the air temperature is 5°F. After one minute the thermometer reads 55°F, and after 5 minutes it reads 30°F. What is the initial temperature of the inside room? dT dT dT =k ( T −T m )=¿ =kdt =¿∫ =∫ kdt dt (T −T m) (T −T m ) ln |T −70|=kt +c=¿T =T m +C ekt Cuando temperatura del aire: 5°F entonces T m=5 ° F T(0) = T 0 T0 = 5 + C

=>

C = T 0−5

T =5+(T 0 −5)e kt Termometro marca 55°F luego de 1 minuto 55=5+(T 0−5)e kt

( T 0−5 ) e t =50 …(1) Termometro marca 55°F luego de 5 minutos

30=5+ ( T 0−5 ) e kt

( T 0−5 ) e 5 t =25 …(2) Dividiendo (2) entre (1): e 5 k−k =

25 =¿ k =−0.17329 50

T 0=64.46 ° F

4. A small metal bar. A small metal bar, whose initial temperature was 20°C, is dropped into a large container of boiling water. How long will it take the bar to reach 90°C if it is known that its temperature increases 2°C in one second? How long will it take the bar to reach 98°C?

dT dT dT =k ( T −T m )=¿ =kdt=¿∫ =∫ kdt dt (T −T m) (T −T m ) ln |T −70|=kt +c=¿T =T m +C ekt

Como el agua en el contenedor esta hierviendo: T m=100 ° C

T(0) = 20 ° C 20 = 100 + C

=>

C = 20−100

T =100−80 e kt

La temperature aumenta en 2°C en 1 segundo:

20+2=100−80 e k 80 e k =78 k =−0.02532=¿T =100−80 e−0.02532t

Cuando la temperature alcanza 90°C:

90=100−80 e−0.02532 t=¿−0.02532 t=ln

( 1080 )

T = 82.13s

Cuando la temperature alcanza 98°C: 98=100−80 e−0.02532 t=¿−0.02532 t=ln

( 802 )

T = 145.69s

5. Thermometer. A thermometer reading 70°F is placed in an oven preheated to a constant temperature. Through a glass window in the oven door, an observer records that the thermometer reads 110°F after 1/2 minute and 145°F after 1 minute. How hot is the oven?

dT dT dT =k ( T −T m )=¿ =kdt =¿∫ =∫ kdt dt (T −T m) (T −T m ) ln |T −70|=kt +c=¿T =T m +C ekt

T(0) = 70°F C=70−T m T =T m +(70−T m)e kt

Cuando el termometro marca 110°F luego de ½ minute

110=T m+ ( 70−T m ) e 0.5t =¿ 110−T m =( 70−T m ) e 0.5k … ( 1 )

Cuando el termometro marca 145°F luego de ½ minute

145=T m + ( 70−T m ) e k =¿ 145−T m=( 70−T m ) e k … (2)

Dividiendo (2) entre (1):

e 0.5 k =

145−T m … (3) 110−T m

Sistema de ecuaciones entre (3) y (1)

145−T m 110−T m = 110−T m 70−T m (145−T m )(70−T m ¿=(110−T m)2 10150−215T m +T m2=12100−220T m +T m2 10150−215T m=12100−220 T m 5 T m=1950=¿T m= T m=390 ° F

1950 5