Termodinamica

PROBLEMAS RESUELTOS DEL TEMA 2. TEMPERATURA Y CALOR Nota: La sección 17 se refiere al Capítulo del Libro de Sears Sección 17.2 Termómetros y escalas de temperatura 17.1 Convierta las siguientes temperaturas Celsius a Fahrenheit: a) -62.8°C. La temperatura mas baja registrada en Norteamérica (3 de febrero de 1947, Snag Yukon); b) 56.7°C, la temperatura mas alta registrada en EE.UU. (10 de julio de 1913, Death Valley, California); c) 31.1°C, la temperatura media anual mas alta de l mundo (Lugo Ferrandi, Somalia). 17.1: Ec. (17.1), a) (9 5)( - 62.8) + 32 = -81.0°F. b) (9 5)(56.7 ) + 32 = 134.1°F.

c) (9 5)(31.1) + 32 = 88.0°F. 17.2 Calcule las temperaturas Celsius que invierno en Seatle (41°F); b) un caluroso c) un frió día de invierno en el norte de Ec. (17.2), a) (5 9)(41.0 - 32 ) = 5.0°C. 17.2: c) (5 / 9 )(- 18 - 32 ) = -27.8°C.

corresponden a: a) una noche de día de verano en Palm Springs (107.0°F) Manitota ( 18°F) b) (5 9 )(107 - 32 ) = 41.7°C.

17.3 Imagine que trabaja en un laboratorio de prueba de materiales y su jefe le dice que aumente la temperatura de una muestra en 40°C. El único termómetro que encuentra en su mesa de trabajo esta graduado en °F. Si la temperatura inicial de la inicial de la muestra es de 68.2°F, ¿Qué temperatura deberá tener en °F una vez que se haya efectuado el aumento pedido? 17.3: 1 C° = 95 F°, so 40.0 = 72.0 F° T2 = T1 + 70.0 F° = 140.2°F 17.4 El 22 de enero de 1943, la temperatura en Spearfish, Dakota del Sur, subió de 4°F a 45°F en solo 2 minutos. Calcule el cambio de temperatura en grados Celsius. b) La temperatura en Browing, Montana, era de 44°F el 23 de enero de 1916. Al dia siguiente la temperatura cayó a 56.0°F. Calcule el cambio en grados Celsius. 17.4: a) (5 9) (45.0 - (-4.0)) = 27.2° C. b) (5 9) (-56.0 - 44) = -55.6° C. 17.5 a) Imagine que se siente mal y le dicen que tiene una temperatura de 40.2°C ¿Qué temperatura tiene en °F? ¿Debe preocuparse? b) El informe matutino de Sydney cita una temperatura de 12°C ¿Cuánto es esto en °F? 17.5: a) Ec. (17.1), (9 5)(40.2) + 32 = 104.4°F, Es motivo de preocupacion. b) (9 5)(12 ) + 32 = 53.6°F, or 54°F .

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17.6 Un ªblue northerº pasa por Lubbock, Texas; una tarde de septiembre y la temperatura baja11.8°C en una hora. Calcule el cambio de temperatura en °F. (9 5)(11.8) = 21.2 F° 17.6: 17.7 Dos vasos de agua, A y B, están inicialmente a la misma temperatura. La temperatura del agua del vaso A se aumenta 10°F y la del vaso b, 10 K. ¿Cuál vaso esta ahora a mayor temperatura? Explique. 17.7: 1 K = 1 C° = 95 F° ,. por lo que un aumento de la temperatura de 10 K correspo nde a un aumento de 18 F°. El vaso B tiene la temperatura más alta 17.8 Se coloca una botella de refresco en un refrigerador y se deja ahí hasta que su temperatura ha bajado 10.0K. Calcule el cambio de temperatura en: a) °F y b) °C 17.8: (b), ΔTC = ΔTK = -10.0 C°. (a), ΔTF = 95 ΔTC = 95 (- 10.0 C°) = -18.0 F°. Sección 17.3 Termómetros de gas y escala Kelvin 17.9 Convierta las siguientes temperaturas record a la escala Kelvin: a) La temperatura más baja registrada en los 48 estados contiguos de EE.UU. ( 70°F en Rogers Pass, Montana, el 20 de enero de 1954); b) la temperatura mas alta en Australia (127°F en Cloncurry, Queensland, el 16 de enero de 1889) c) la temperatura mas baja registrada en el hemisferio norte ( 90°F en Verkhoyansk, Liberia en 1892). 17.9: Combinando Ec. (17.2) y Ec. (17.3), 5 TK = (TF - 32°) + 273.15, 9 Y la sustitución de las temperaturas Fahrenheit dada, da a) 216.5 K, b) 325.9 K, c ) 205.4 K. 17.10 Convierta las siguientes temperaturas Kelvin a las escalas Celsius y Fahrenheit: a) la temperatura al medio día en la superficie de la luna (400°K); b) l a temperatura en la parte alta de las nubes de la atmósfera de Saturno (95K); c) la temperatura en el centro del Sol (1.55 x 107 K) 17.10: (En estos cálculos, las cifras extra se guardaban en los cálculos intermedios  para llegar a los resultados numéricos.) a) TC = 400 - 273.15 = 127°C, TF = (9 / 5)(126.85) + 32 = 260°F. b) TC = 95 - 273.15 = -178°C, TF = (9 / 5)(-178.15) + 32 = -289°F. c) TC = 1.55 × 107 - 273.15 = 1.55 × 107°C, TF = (9 / 5)(1.55 × 107 ) + 32 = 2.79 × 107°F.

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17.11 El punto de ebullición normal del neon líquido es 245.92°C. Exprese esta temperatura en la escala Kelvin. 17.11: Ec. (17.3), TK = (-245.92°C) + 273.15 = 27.23 K. 17.12 La relación de las presiones de un gas en el punto de fusión del platino y en el punto triple del agua, manteniendo el volumen del gas constante, es 7.476. ¿A que temperatura Celsius se funde el platino? 17.12 Ec. (17.4), (7.476)(273.16 K) = 2042.14 K - 273.15 = 1769°C. 17.13 Un terremoto de gas registro una presión absoluta correspondiente a 325mm de mercurio, estando en contacto con agua en el punto triple. ¿Qué presión indicara en contacto con agua en el punto de ebullición normal? .15 K ) = 444 mm. 17.13: Ec. (17.4), (325.0 mm)( 373 273.16 K 17.14 Al igual que la escala Kelvin, la escala Ranking es una escala absoluta de temperatura: el cero absoluto es cero grados Ranking (0°R). Sin embargo, las unidades de esta escala tienen el mismo tamaño que las de la escala Fahrenheit, no las de la escala Celsius. De el valor numérico de la temperatura del punto trip le del agua en la escala Rankine. 17.14: En la escala Kelvin, el punto triple es 273,16 K, de modo °R = (9/5)273.15 K = 491.69°R. que también podría buscar en la Figura 17.7 y nota que la escala Fahrenheit - 460°F to + 32°F y se extiende desde la conclusión de que el punto triple es de unos 492 17.15 Termómetro de gas de volumen constante. Usando un termómetro de gas, un experimentador determino que la presión en el punto triple del agua (0.01°C) era de 4.80 x 104 Pa, y en el punto de ebullición normal del agua (100°C), 6.50 x 104 Pa. a) suponiendo que la presión varia linealmente con la temperatura, use estos datos para calcular la temperatura Celsius en la que la presión del gas seri a cero ( es decir obtenga la temperatura Celsius del cero absoluto). b) ¿el gas de este termómetro obedece con precisión la ecuación (17.4)? Si así fuera y la presión a 100°C fuera 6.50 x 104 Pa, ¿Qué presión habría medido el experimentador a 0.01°C? (como veremos la sección 18.1, la ecuación (17.4) solo es exacta para gases a muy baja densidad.) 17.15: Desde el punto pendiente la fórmula para una línea recta (o de regresión lineal  que, aunque tal vez no proceda, puede ser conveniente para algunas 4.80 × 10 4 Pa calculadoras), (0.01°C) - (100.0°C) = -282.33°C, 6.50 × 10 4 Pa - 4.80 × 10 4 Pa Que es - 282°C para las tres b) Equation (17.4) was not obeyed precisely. If it were, the pressure at the tri ple point ×10 4 Pa would be P = (273.16) 6.50373 = 4.76 × 10 4 Pa. .15 ( )

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Sección 17.4 Expansión térmica 17.16 Fricción del aire y expansión térmica. El avión supersónico Concorde (hecho principalmente de aluminio) tiene 62.1 m de longitud en la pista en un día ordinar io (15°C). Volando al doble de la rapidez del sonido, la fricción con el aire calienta la superficie del Concorde y alarga el avión 25cm. (La cabina de pasajeros esta en rodillos; el avión se expande a su alrededor) ¿Qué temperatura tiene la superficie del Concorde en vuelo? -1 17.16: ΔT = (ΔL ) (aL0 ) = 25 × 10-2 m 2.4 × 10-5 (C°) (62.1 m = 168° C, Por lo que l  temper tur  es 183° C . ( ) (( )) 17.17 El puente Humber de Ingl terr  tiene el cl ro individu l m s l rgo del mundo (1140 m). C lcule el c mbio de longitud de l  cubiert  de cero del cl ro si l  temper tur ument  de 5°C  18.0°C. 17.17: aL0 ΔT = (1.2 × 10-5 (C°) -1 )(1410 m)(18.0° C - (-5.0)°C) = +0.39 m.

17.18 Ajuste Estrecho. Los remaches de aluminio para construcción de aviones se fabrican un poco mas grande que sus agujeros y se enfrían con hielo seco (CO2 solidó) antes de insertarse. Si el diámetro de un agujero es de 4.500 mm ¿Qué diámetro debe tener un remache a 23°C para que su diámetro sea igual al del agujero cuando se enfría a 78°C, la temperatura del hielo seco? Suponga que el coeficiente de expansión es constante, con el valor dado en la tabla 17.1. 17.18: d + Δd = d (1 + aΔT ) = (0.4500 cm)(1 + (2.4 × 10-5 (C°) -1 )(23.0° C - (-78.0°C))) = 0.4511 cm = 4.511 mm. 17.19 Un centavo de dólar tiene 1.9000 cm. de diámetro a 20.0°C, y esta hecho de una aleación (principalmente zinc) con un coeficiente de expansión lineal de 2.6 x 10 5 K 1. ¿Qué diámetro tendría: a) en un día caluroso en Death Valley (48°C)? b) ¿en una noche fría en las montañas de Groelandia ( 53.0°C)? 17.19: a) aD0 ΔT = (2.6 × 10 -5 (C°) -1 )(1.90 cm) (28.0°C) = 1.4 × 10-3 cm, entonces el diámetro es 1.9014 cm. b) aD0 ΔT = -3.6 × 10 -3 cm, el diámetro es 1.8964 cm. 17.20 La varilla del péndulo de un reloj es de latón. Calcule su cambio fraccionario de longitud si se enfría a 19.50°C a 5.00°C 17.20: aΔT = (2.0 × 10-5 (C°) -1 )(5.00° C - 19.5° C) = -2.9 × 10-4.

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17.21 Una varilla metálica tiene 40.125 cm. de longitud a 20.0°C y 40.148 cm. a 45.0°C. Calcule el coeficiente medio de expansión lineal para la varilla en este intervalo de temperatura. 17.21: a = (ΔL) ( L0 ΔT ) = 2.3 × 10-4 m 40.125 × 10-2 m (25.0 C°) = 2.3 × 10 -5 (C°) ( -1 ) (( ) ) . 17.22 Un cilindro de cobre esta a 20.0°C ¿a que temperatura aumentara su volumen 0.150%? -3 ΔV V 17.22: Ec. (17.8), ΔT = b 0 = 5.11.×5010×-105 K -1 = 29.4°C, so T = 49.4°C. 17.23 Un tanque su terráneo con capacidad de 1,700 L (1.70m3) se llena con etanol a 19.0°C. Una vez que el etanol se enfría a la temperatura del tanque y el suelo, que es 10.0°C, ¿Cuánto espacio de aire ha rá so re el etanol del tanque? (suponga que el volumen del tanque no cam ia) -1 17.23 b V0 ΔT = 75 × 10-5 (C°) (1700 L )(- 9.0°C ) = -11 L, por lo que son 11 L de aire.

( ) 17.24 Un tanque de acero se llena totalmente con 2.80m3 de etanol cuando ambos el tanque con el etanol están a 32.0°C. Una vez que el tanque y el contenido se hayan enfriado a 18.0°C, ¿Qué volumen adicional de etanol podrá meterse al tanque? 17.24: El cambio de temperatura es ΔT = 18.0° C - 32.0° C = -14.0 C°. el volumen de etanol contrae más que el volumen del tanque de acero que, por lo que la suma adic ional de etanol que se pueden poner en el tanque es ΔVsteel - ΔVethanol = ( bsteel - bethanol ( )( )

)V0

ΔT

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e derraman 8.95 cm3 de mercurio. El coeficiente de expansión de volumen ( b)

del mercurio es de 18.0 x 10 5 K 1; calcule el coeficiente de expansión de volumen del vidrio. 17.25: La cantidad de mercurio que des orda es la diferencia entre la variación de l volumen del mercurio y de que el vaso; 8.95 cm 3 -1 b Vaso = 18.0 × 10 -5 K -1 = 1.7 × 10 -5 (C°) . 3 1000 cm (55.0°C ) ( ( ) ) 17.26 a) Si un área medida en la superficie de un cuerpo solidó es A0 a cierta temperatura inicial y cam ia en ΔA cuando la temperatura cambia en Δ7, demuestre que ΔA = (2 a)A0ΔT. Donde a es el coeficiente de exp nsión line l. b) un  l min  circul r de luminio tiene 55.0 cm. de diámetro  15.0°C. ¿Cuánto

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c mbi  el áre  de un  c r  de l  lámin  cu ndo l  temper tur 27.5°C? 17.26: ) A = L2 , ΔA = 2 LΔL = 2 ΔLL L2 = 2 ΔLL A0 . But ΔLL ΔA = 2aΔTA0 = (2a )A0 ΔT .

ument

= aΔT , and so

b)

ΔA = (2a

)Ao

ΔT = (2 ) (2.4 × 10-5 (C°) -1 ) ( p

× (.275 m ) ) (12.5°C ) = 1.4 × 10-4 m 2 .

2 17.27 Un o erario hace un agujero de 1.350 cm. de diámetro en una laca de acero a 25°C. ¿Qué área transversal tendrá el agujero: a) 25°C; b) si la laca se calienta a 175°C? Su onga que el coeficiente de ex ansión lineal es constante dentro de este intervalo (sugerencia: véase el ejercicio 17.26) pD 2 p 2 = (1.350 cm ) = 1.431 cm 2 . 17.27: a) A0 = 4 4 b) A = A0 (1 + 2 a ΔT ) = 1.431 cm 2 1 + (2 ) 1.20 × 10-5°C (150°C ) = 1.437 cm 2 .

17.28 Imagine que acaba de comenzar a trabajar como ingeniero mecánico en Motores, S.A. y le encargaron diseñar pistones de latón que se deslizaran dentro de cilindros de acero. Los motores en los que se usaran los pistones operaran a temperaturas entre 30°C y 150°C. Suponga que los coeficientes de expansión son constantes dentro de ese intervalo de temperatura. a) si el pistón apenas cabe dentro del cilindro a 20°C, ¿los motores podrán operar a temperaturas más altas? Explique, b) si los pistones cilíndricos tienen un diámetro de 25.00 cm. a 20°C, ¿Qué diámetro mínimo deberá tener los cilindros a esa temperatura para que los pistones operen a 150°C? 17.28: (a) No, el latón se expande más que el acero. (b) El diámetro interior del cilindro en acero en 20°C At 150°C : DST = DBR ( )( ( ) )

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17.29 Las marcas de una regla de aluminio y una de latón están perfectamente aliadas a 0°C. ¿Qué separación habrá entre las marcas de 20 cm. de las dos reglas a 100°C, si se mantiene una alineación precisa de los extremos izquierdos de las reglas? 17.29: La regla de aluminio se expande a una nueva longitud de La regla de latón se expande a una nueva longitud de L = L0 (1 + aΔT ) = (20.0 cm)[1 + (2.0 × 10-5 (C°) -1 )(100 C°)] = 20.040 cm

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La sección de la regla de aluminio será más largo por 0.008 cm. 17.30 Una varilla de latón de 185 cm. de longitud y de 1.60cm de diámetro. ¿Qué fuerza debe aplicarse a cada extremo para impedir que se contraiga al enfriarse de 120°C a 10°C? 17.30: Ec. (17.12), F = -YaΔTA = -(0.9 × 1011 Pa)(2.0 × 10- 5 (C°) -1 )(-110°C)(2.01 × 10- 4 m 2 ) = 4.0 × 104 N. 17.31 a) un alambre con longitud de 1.50 m a 20.0°C se alarga 1.9 cm. al calentarse a 420°C. Calcule su coeficiente medio de expansión lineal para este intervalo de temperatura, b) el alambre se tiende sin tensión a 420°C. Calcule el esfuerzo en el si se enfría a 20°C sin permitir que se contraiga. El modulo de Young del alambre es 2.0 x 1011 Pa.17.31: a) -2 -5 -1 a = (ΔL L0 ΔT ) = (1.9 × 10 m) ((1.50 m)(400 C°) ) = 3.2 × 10 (C°) . b) YaΔT = YΔ L L0 = (2.0 × 1011 Pa)(1.9 × 10-2 m) (1.50 m) = 2.5 × 109 Pa. 17.32 Rieles de acero para un tren se tienden en segmentos de 12.0 m de longitud colocados a tope en un día de invierno en que la temperatura es de 2°C, a) ¿Cuánto espacio debe dejarse entre los rieles adyacentes para que apenas se toquen en verano cuando la temperatura suba a 33°C? b) si los rieles se tienden en contacto ¿a que esfuerzo se someterán un día de verano n el que la temperatura se a de 33°C? 17.32: a) ΔL = aΔTL = (1.2 × 10-5 K -1 )(35.0 K)(12.0 m) = 5.0 × 10 -3 m. b)Usando valores absolutos en Ec. (17.12), F = YaΔT = (2.0 × 1011 Pa)(1.2 × 10-5 K -1 )(35.0 K) = 8.4 × 107 Pa. A

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17.34 Al correr un estudiante de 70 kg genera energía térmica a razón de 1,200 watts, para mantener una temperatura corporal constante de 37°C, esta energía debe eliminarse por sudor u oreos mecanismos. Si tales mecanismos fallaran y no pudiera salir calor del cuerpo, ¿cuanto tiempo podría correr el estudiante antes de sufrir un daño irreversible? (las estructuras proteinitas del cuerpo se dañan irreversiblemente a 44°C o mas. La capacidad calorífica especifica del cuerpo humano es de alrededor de 3,480 J/Kg. · k poco menos que la del agua; la

diferencia se debe a la presencia de: proteínas grasas y minerales, cuyo calor especifico es menor que el del agua). 17.34: t = Q mcΔT (70 kg)(3480 J kg

× K )(7 C°)

= = = 1.4  103 s, alrededor de 24 min. P P (1200 W ) 17.35 Al pintar la punta de una antena de 225 metros de altura, un trabajador de ja caer sin querer una botella de agua de 1.00 L de su ponchera. La botella cae en unos arbustos en el suelo y no se rompe. Si una cantidad de calor igual a la magnitud del cambio de energía mecánica del agua pasa al agua. ¿Cuánto aumentara su temperatura? 17.35: UsandoQ=mgh in Ec. (17.13) y para la solucion ΔT da gh (9.80 m s 2 )(225 m)

ΔT = = = 0.53 C°. c (4190 J kg.K ) 17.36 Una caja con fruta de 50.0 Kg. y calor especifico de 3,650 j/Kg. k baja deslizándose por una rampa de 8.00 m de longitud inclinada 36.9° bajo la horizontal. A) si la caja estaba en reposo arriba de la rampa y tiene una rapide z de 2.50 m/s, ¿Cuánto trabajo efectuó la fricción sobre ella? B) si una cantidad de calor igual a la de dicho trabajo pasa a la fruta y esta alcanza una temperatura final uniforme, ¿Qué magnitud tiene el cambio de la temperatura? 17.36: a) El trabajo realizado por la fricción es la pérdida de energía mecánica, 1 1

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móvil contiene u 1.60 Kg. de aluminio y 0.30 Kg. de hierro, y esta diseñada ara o erar a 210°C ¿Cuánto calor se requiere ara elevar su tem eratura de 20°C a 210°C? 17.37: (210° C - 20°C )((1.60 kg )(910 J kg × K ) + (0.30 kg )(470 J kg × K )) = 3.03 J.

 105

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17.38 Un clavo que se clava en una tabla sufre un aumento de temperatura. Si suponemos que el 60% de la energía cinética de un martillo de 1.80 Kg. que se mueve a 7.80 m/s se transforma en calor que fluye hacia el clavo y no sale de el , ¿Cuánto aumentara la temperatura de un clavo de aluminio de 8.00 g golpeado 10 veces? 17.38: Asumiendo Q = (0.60 )  10  K , 1 (6) 12 (1.80 kg )(7.80 m s ) = 45.1 C°. MV 2 K

ΔT = (0.60) × 10 × =62 = 8.00 × 10- 3 kg (910 J kg

× K )

mc mc 2 ( ) 17.39 Una tetera de aluminio de 1.50 kg que contiene 1.80 Kg. de agua se pone en la estufa. Si no se pierde calor al entorno ¿cuanto calor debe agregarse para aumentar la temperatura de 20.0°C a 85.0°C? 17.39: (85.0° C - 20.0° C )((1.50 kg )(910 J kg × K ) + (1.80 kg )(4190 J kg × K )) = 5.79

 10 5 J.

17.40 tratando de mantenerse despierto para estudiar toda la noche un estudiante prepara una taza de café colocando una resistencia eléctrica de inmersión de 200

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líquido? Explique. (120 s )(65.0 W ) Q 17.41: a) c = = = 2.51  103 J kg × K. mΔT (0.780 kg )(22.54°

C - 18.55° C ) b) Un estimado, el calor Q es en realidad inferior a la potencia en veces el int ervalo de tiempo 17.42 Imagine que le dan una muestra de metal y le piden determinar su calor específico. Pesa la muestra y obtiene un valor de 28.4 N. añade con mucho

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cuidado 1.25 x 10 J de energía calorífica la muestra y observa que su temperatura aumenta 18.0°C ¿Qué calor especifico tiene la muestra? 17.42: El cambio de temperatura es ΔT = 18.0 K, so c= ( )( ) Q gQ 9.80 m s 2 1.25 × 104 J = = = 240 J kg × K. (28.4 N )(18.0 K ) mΔT w ΔT

17.43 Se añaden 8,950 J de calor a 3.00 moles de hierro. A) determine el aumento de temperatura del hierro. B) si añaden la misma cantidad de calor a 3.00 Kg. de hierro ¿Cuánto subirá su temperatura? C) Compare los resultados de las partes a y b y explique la diferencia. Sección 17.6 Calorimetría y cambios de fase 17.43: a) Q = mcΔT , c = 470 J kg × K Tenemos que encontrar la masa de 3.00 mol: ( )

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= (0.50 kg )(15 C°) mΔT

c= 17.45 Un trozo de 500g de un metal desconocido, que ha estado en agua hirviente durante varios minutos, se deja caer rápidamente en una vaso de espuma de poliestireno aislante que contiene 1.00 Kg. de agua a temperatura ambiente (20°C). Después de esperar y agitar suavemente durante 5 minutos se observa que la temperatura del agua ha alcanzado un valor constante de 22.0°C

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a) suponiendo que el baso absorbe una cantidad despreciable de calor y que no se pierde calor al entorno ¿Qué calor especifico tiene el metal? B) ¿Qué es mas útil para almacenar calor, este metal o un peso igual de agua? Explique c) suponga que el calor absorbido por el vaso no es despreciable ¿Qué tipo de error tendría el calor especifico calculado en la parte a (seria demasiado grande, demasiado pequeño o correcto)? Explique. 17.45: a) Qwater + Qmetal = 0 m water c water ΔTwater + m metal c metal ΔTmetal = 0 (1.00kg )(4190 J kg × K )(2.0 C°) + (0.500 kg )(cmetal )()( - 78.0 C°) = 0 cmetal = 215 J/kg × K b) El agua tiene una mayor capacidad de calor específico a fin de almacenar más calor por grado de los cambios de temperatura. c) Si algo de calor entró en la espuma de poliestireno entonces realmente debe ser mayor que en la parte (a), por lo que el valor real es más grande que el valor cal culado que sería más pequeño que el valor real. 17.46 Antes de someterse a un examen medico anual, un hombre de 70.00 Kg. cuya temperatura corporal es de 37°C consume una lata entera de 0.355 L de gaseosa (principalmente agua) que esta a 12°C a) determine su temperatura corporal una vez alcanzado el equilibrio. Desprecie cualquier calentamiento por el metabolismo del hombre. El calor especifico del cuerpo del hombre es de 3,480 J ·

k. b) ¿el cambio en su temperatura corporal es lo bastante grande como para medirse con un termómetro medico? 17.46: a) Que el hombre designado por el subíndice m y el "agua" por w, y T es la temperatura final de equilibrio. - m m C m ΔTm = m w C w ΔTw - mmCm (T - Tm ) = mw Cw (T - Tw ) m m C m (Tm - T ) = m w C w (T - Tw )

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vuelva a 37.0°C si su cuerpo desprende energía a razón de 7.00 x 10 kJ/dia (la tasa metabólica basal TMB) ¿Cuánto tardara en hacerlo? Suponga que toda la energía desprendida se invierte en elevar la temperatura.

Δt 17.47La tasa de pérdida de calor s ΔQ Δt. ΔΔQt = mCΔΔt T , or Δt = mC ΔQ . Interesting numbers, ( ) ( ) Δt = ( 70.355 kg)(3480 J kg.° C)(0.15° C) 7×10 6 J day Δt = 0.005 d, or Δt = 7.6 minutos. Esto cuenta para las madres,

tomar la temperatura de un niño enfermo varios minutos después de que el niño tomo alg o para beber.

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b) mLf P = ( 0.550 kg)(334 10 3 J kg ) (800 J min ) = 230 min, so the time until the ice has melted is 21.7 min + 230 min = 252 min. 17.51 La capacidad de los acondicionadores de aire comerciales a veces se expresa en ªtoneladasº las toneladas inglesas de hielo (1 ton=2,00 lb.) que la unidad puede congelar a partir de agua a 0°C en 24h. Exprese la capacidad de un acondicionador de aire de una tonelada en btu/h y en watts.

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((4000 lb) 17.51: 2.205 lb kg ) )(334  103 J kg) = 7.01 kW = 2.40  104 Btu hr. (86,400 s) 17.52 Quemaduras de vapor vs. quemaduras de agua. ¿Cuánto calor entra en su piel si recibe el calor liberado por a) 25 g de vapor de agua que inicialmente e sta a 100°C, al enfriarse a la temperatura de la piel (34°C)? b) 25 g de agua que inicialmente esta a 100°C al enfriarse a 34.0°C? c) ¿Qué le dice esto acerca de la severidad relativa de las quemaduras con vapor y con agua caliente? 17.52: a) m(c ΔT + Lv ) = (25.0 × 10-3 kg) (4190 J kg × K)(66.0 K) + 2256  103 J kg = ( -3 )

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intervalo de 1-2°C, un camello deshidratado deja que su temperatura corporal baje a 34°C de noche y suba a 40°C de día. Para ver lo eficaz que es este mecanismo para ahorrar agua, calcule cuantos litros de agua tendría que beber un camello de 400 Kg. si tratara de mantener su temperatura corporal en 34°C mediante evaporación de sudor durante el día (12 h) en lugar de dejar que suba a 40°C (la

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capacidad calorífica especifica de un camello u otro mamífero es la de una persona representativa, 3,480 J/kg · k. el calos de vaporización del agua a

de 2.42 x 10 J/Kg.). 17.55: La masa de agua que el camello guarda McΔT (400 kg)(3480 J kg × K)(6.0 K)

34°C es

= = 3.45 kg, Lv (2.42  106 J kg) es un volumen de 3.45 L. 17.56 en un experimento de laboratorio de física un estudiante sumergió 200 centavos (cada uno con mas de 3.00 g) en agua hirviendo. Una vez alcanzado el equilibrio térmico los saco y los puso en 0.240 kg de agua a 20°C en un recipiente aislado con masa despreciable. Calcule la temperatura final de las monedas (hechas con una aleación de zinc con una capacidad calorífica de 390 J/Kg · k)

17.56:

Para este caso, el álgebra se reduce a ⎛ ((200)(3.00 × 10-3 kg ))(390 J kg × K ⎜ ⎟

)(100.0 C°)

⎞

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agua, el calor y la capacidad de la muestra es ((0.200 kg)(4190 J kg × K ) + (0.150 kg)(390 Q c= = (0.0850 kg)(73.9 C°) mΔT = 1010 J kg × K,

J kg × K ))(7.1 C°)

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o 1000 J kg × K a las dos cifras a la que el cambio de temperatura que se conoce. 17.59 Un vaso asilado con una masa despreciable contiene 0.250 kg de agua a 75°C ¿Cuántos kg de hielo a -20°C debe ponerse en el agua para que la temperatura final del sistema sea 30°C? 17.59: El Ccalor perdido por el agua - Q = (0.250 kg )(4190 J kg × K )(45.0 C°) = 4.714  104 J, y la masa de hielo que se necesita -Q mice = cice ΔTice + Lf + c water ΔTwater

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mP b c Pb (200°C -  ) = (m w + mice )c w  + mcu ccu  + mice Lf . Resolución de la tem eratura final  y la utilización de valores numéricos ⎛ (0.750 kg)(130 J kg × K)(255 C°) ⎞

⎜ ⎟ ⎜ - (0.018 kg)(334 × 103 J kg) ⎟⎠ ⎝ T= = 21.4°C. ⎛ (0.750 kg)(130 × J kgK)

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17.65: m c)100.0°C - (222 K m)(12.00 × 10-2 m) = 73.3°C.

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17.66 Un extremo de una varilla metálica aislada se mantiene a 100 ºC y el otro se mantiene a 0 ºC con una mezcla de hielo agua. La varilla tiene 60.0 cm. de longitud de área transversal de 1.25 cm². El calor conducido or varilla funde 8.50 g hielo en 10.00 min. Calcule la conductividad térmica k del metal. dm Y resolviendo Ec. (17.21) for k, 17.66: Usando la regla de la cadena, H = dQ dt = Lf dt dm L k = Lf dt AΔT

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