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Alex777 [14]
3 years ago
13

An acceleration of a freely falling body (9.8m/s^2) to ft/s^2

Physics
1 answer:
Ronch [10]3 years ago
5 0
9.8 m/s^2 = 32.2 ft/s^2

hope this helps :)
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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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Answer:

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

Explanation:

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

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Donde:

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T - Temperatura del sistema, en Kelvin.

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Donde:

m - Masa del sistema, en kilogramos.

L_{v} - Calor latente de vaporización, en joules

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

Como vemos, el cambio de la entropía asociada al cambio de fase del mercurio es directamente proporcional a la masa del sistema. Si tenemos que m = 0.50\,kg,L_{v} = 2.7\times 10^{5}\,\frac{J}{kg} and T = 630.15\,K, entonces el cambio de entropía es:

\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}

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

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