La razón principal del alto costo de la destilación como...

La razón principal del alto costo de la destilación como método para purificar agua es la gran cantidad de energía que se necesita para calentar y vaporizar el agua. (a) Con base en la densidad, el calor específico y el calor de vaporización del agua del apéndice B, calcule la cantidad de energia necesaria para evaporar 1 galón de agua que inicialmente está a $20^{\circ} \mathrm{C}$. (b) Si la energía se obtiene de electricidad que cuesta 0.085 dólares por kwh, calcule su costo. (c) Si el agua destilada se vende en las tiendas a 1.26 dólares por galón, ¿qué porcentaje del precio de venta representa el costo de la energia?

La razón principal del alto costo de la destilación como método para purificar agua es la gran cantidad de energía que se necesita para calentar y vaporizar el agua.
(a) Con base en la densidad, el calor específico y el calor de vaporización del agua del apéndice B, calcule la cantidad de energia necesaria para evaporar 1 galón de agua que inicialmente está a $20^{\circ} \mathrm{C}$. (b) Si la energía se obtiene de electricidad que cuesta 0.085 dólares por kwh, calcule su costo. (c) Si el agua destilada se vende en las tiendas a 1.26 dólares por galón, ¿qué porcentaje del precio de venta representa el costo de la energia?

Key Concepts

Latent Heat of Vaporization

Latent heat of vaporization is the energy required per unit mass to change a substance from a liquid to a vapor without altering its temperature. This concept is essential in processes such as distillation because it quantifies the significant energy needed for the phase change after the substance has been heated to its boiling point.

Specific Heat Capacity

Specific heat capacity is a property of a substance that indicates the amount of energy required to raise the temperature of a unit mass by one degree Celsius. This is an important concept when calculating the energy needed to heat a substance from its initial temperature to the boiling point before vaporization occurs.

Energy Conversion

Energy conversion involves translating energy values from one unit to another, such as converting joules to kilowatt-hours. This concept is key when linking calculated energy requirements to practical measures like cost, especially in industrial or laboratory settings.

Cost Analysis of Energy Consumption

Cost analysis of energy consumption connects the amount of energy used in a process to its economic cost by applying the energy price per unit. This is fundamental when evaluating the feasibility and expense of energy-intensive processes, ensuring that operational decisions are informed by cost calculations.

Unit Conversion

Unit conversion is the process of translating measurements from one unit system to another, such as converting volume from gallons to liters or mass from one scale to another. This is necessary for applying standard values and physical constants correctly in calculations, ensuring that energy estimations and cost computations are accurately conducted.

Transcript

00:01 Okay, now we do a problem 85.

00:05 And you have to think about this.

00:07 For example, liquid right here, right? liquid.

00:10 So you need that from here 20 degrees, right? and now you know from this to 100 degrees.

00:19 This is the temperature of water boiling.

00:22 And during water evaporates until its end, the temperature always 100 celsius degrees.

00:31 So they need to kill for fusion, and q vaporase.

00:39 So you will have to add q both of them, a, b, and b, c, right? from 20 degrees to 100 celsius degree and from b to c, this is the same temperature, but this is q vaporate of the water.

00:55 And from a to c, that means q for fusion.

01:08 So we have first one.

01:10 Because we got the gallon, right? gallon, so we have to convert gallon to gram because we get all of this for a couple of grams.

01:18 It's not for a gallon, so you have to convert gallon to gram.

01:41 So for this, we have this about 3 .785 times 3 .3 grams, right? and now we got the q fusion.

01:56 This must be equal mcf delta t, right? this is specific heat of water.

02:03 So that 3 .4 -875 times 3 grams.

02:10 And specific hope of the water liquid is cool.

02:17 4 .1 .184 jow over gram c.

02:26 Right.

02:27 And for this one, where temperature equals 100.

02:36 And for this, we will have this about, now, if we continue to convert two kilograms, so you get one kilogram chl -chouk, kilo -jou over 10 negative, 10 -power -3 chow.

02:55 Why i do this because i know the vapor, the energy of vapor, vapor of water is kilo -chow, not chow.

03:05 So because later we have to add them together, so i convert them together right here.

03:10 So for this you have about 1 .267.

03:20 Okay.

03:23 So we here, okay.

03:27 We have to match the note right here.

03:29 Water, the boiling point of water is 100 celsius degree, right? so this is the initial and this is the final...