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NARA [144]
3 years ago
12

Help!! Please!!

Physics
1 answer:
sattari [20]3 years ago
6 0
<span> the answer is 1,500 & 1,700


</span>
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Cuanto cambia la entropía de 0.50 kg de vapor de mercurio [Lv: 2.7 x 10⁵ j/kg ] al calentarse en su punto de ebullición de 357°
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La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

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Por definición de entropía (S), medida en joules por Kelvin, tenemos la siguiente expresión:

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Q - Ganancia de calor, en joules.

T - Temperatura del sistema, en Kelvin.

Ampliamos (1) por la definición de calor latente:

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Donde:

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Puesto que no existe cambio en la temperatura durante el proceso de vaporización, transformamos la expresión diferencial en expresión de diferencia, es decir:

\Delta S = \frac{\Delta m \cdot L_{v}}{T}

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\Delta S = \frac{(0.50\,kg)\cdot \left(2.7\times 10^{5}\,\frac{J}{kg} \right)}{630.15\,K}

\Delta S = 214.235 \,\frac{J}{K}

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