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gavmur [86]
3 years ago
10

Una barra de plata de 335.2 g con una temperatura de 100 ºC se introduce un calorímetro de aluminio de 60 g de masa que contiene

450 g de agua a 23 ºC. Sr agita la mezcla y la temperatura se incrementa hasta 26 ºC. ¿Cuál es el calor especifico de la plata?
Physics
1 answer:
sdas [7]3 years ago
8 0

Respuesta:

0,0560 cal / gºC.

Explicación:

Cantidad de calor; (Q)

Q = mcΔt; Δt = t2 - t1

m = masa, c = capacidad calorífica específica; Δt = cambio de temperatura

c de agua = 1 cal / gºC

c de aluminio = 0,22 cal / gºC

QTotal = Q de agua + Q de aluminio

Q de agua = 450 * 1 * (26 - 23) = 1350 cal

Q de aluminio = 60 * 0.22 * (26 - 23) = 39.6 cal

QTotal = 1350 + 39,6 = 1389,6 cal

Calor perdido = calor ganado

QTotal = calor perdido

- 1389,6 = 335,2 * c * (26 - 100)

-1389,6 = −24804,8 * c

c = 1389,6 / 24804,8

c = 0,056021 cal / gºC.

Capacidad calorífica específica de la plata = 0,0560 cal / gºC.

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