Se mezclan 20 gramos de agua (1cal/g°C) a 40 °C con 15 gramos de alcohol (0,58cal/g°C) a 30 °C. ¿Cuál ha sido la temperatura de equilibrio térmico?

Answers

Answer 1

Answer:

[tex]T_{EQ}=37\°C[/tex]

Explanation:

¡Hola!

En este caso, para los problemas de equilibrio térmico, consideramos que la energía liberada por la sustancia que inicialmente está caliente (agua), es absorbida por la sustancia que inicialmente está fría (alcohol); thus, we can write:

[tex]Q_{agua}=-Q_{alcohol}[/tex]

La cual puede ser escrita en términos de masa, calor specifico y temperaturas:

[tex]m_{agua}C_{agua}(T_{EQ}-T_{agua})=-m_{alcohol}C_{alcohol}(T_{EQ}-T_{alcohol})[/tex]

De este modo, al resolver para la temperature de equilibrio térmico, obtenemos la siguiente expresión:

[tex]T_{EQ}=\frac{m_{agua}C_{agua}T_{agua}+m_{alcohol}C_{alcohol}T_{alcohol}}{m_{agua}C_{agua}+m_{alcohol}C_{alcohol}}[/tex]

Así, al reemplazar los valores en esta, obtenemos:

[tex]T_{EQ}=\frac{(20g)(1cal/g\°C)(40\°C)+(15g)(0.58cal/g\°C)(30\°C)}{(20g)*(1cal/g\°C)+(15g)(0.58cal/g\°C)} \\\\T_{EQ}=37\°C[/tex]

¡Saludos!


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Explanation:

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Answers

Answer:

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Answers

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Answers

Answer:

Please find the definition and example of neutralization reaction below

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Answers

Answer:

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Sources (jic)

https://www.usgs.gov/special-topic/water-science-school/science/evapotranspiration-and-water-cycle

https://www.pthorticulture.com/en/training-center/basics-of-plant-respiration/

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