# Energy Require from water-steam

• mom2maxncoop
In summary, the conversation is about calculating the amount of energy required to change one quarter of 8.0kg of water at 25°C into steam. The equation used was Q=mcΔt + 0.25mL, and the values for the constants were discussed. It was determined that the additional energy needed to convert the water to steam would be calculated using the latent heat of vaporization. The final calculation for Q resulted in 7120000J.
mom2maxncoop

## Homework Statement

How much energy is required to change one quarter of 8.0kg of water at 25°C into steam.

## Homework Equations

I get confused to exact what equation to use to figure out the energy required.

I *think* it would be this one:
Q$_{t}$=M$_{water}$C$_{water}$$\Delta$T$_{water}$ + .25(M$_{water}$)L$_{f}$

## The Attempt at a Solution

When I do the math, I come up with a ridulous number that makes no sense. I try and try to understand but I can't.

:/.

What are the values of the constants are you using? What value are you getting?

Would it be:

Q=mc$\Delta$t
Q=(8.0kg)(4200J/(kg·°C))(100°C-25°C)
Q=2 520 000

Then since your finding a a quater of it would you then muliply by .25

2520000x.25

=630000J

It seems to easy since the question is worth 5 marks, it feels like I'm missing a step or something :/

It boils down to (pun!) whether you are expected to raise all 8.0 kg of water to 100° and then transform 1/4 of it to steam, or whether you are to take 1/4 of 8.0 kg of water and raise its temperature to 100° and then transform it to steam.

Your value for Q looks okay for raising the 8.0 kg of water to 100°. How much additional energy will it take to convert 1/4 of that water at 100° to steam at 100°?

I loved the pun, I actually had a giggle with that one!

Right now I'll have to find the mass of the steam itself!
Q=

To find the additional energy would I use the equation that I initially posted (would I use latent fusion or Latent vaporiztion)?

BUT I don't know the mass of the steam nor do I know the Energy, so I have NO Idea where to go next :(:(

(it sucks trying to teach yourself everything from Grade 11 math when you've been out of high school for 6 years :/)

The mass is just 1/4 of 8.0 kg. You want to vaporize this mass of water, so you'd use the latent heat of ______?

Vaporization since you want to vaporize the mass.

So with that being said

Q=(8.0kg)(4200J(kg·°C)(100°C-25°C)+.25(8.0kg)(2300000)
Q=2520000 + 4600000
Q=7120000J

Looks good!

Thank you so much!

## What is "energy requirement from water-steam"?

"Energy requirement from water-steam" refers to the amount of energy needed to heat water and convert it into steam. This energy is usually measured in units of joules or calories.

## What factors affect the energy requirement from water-steam?

The energy requirement from water-steam can be influenced by several factors, including the initial temperature of the water, the pressure at which the steam is produced, and the quantity of water being heated.

## How is the energy requirement from water-steam calculated?

The energy requirement from water-steam can be calculated using the formula: Q = m*C*ΔT, where Q is the energy in joules, m is the mass of water in kilograms, C is the specific heat capacity of water (4.186 J/g°C), and ΔT is the change in temperature in degrees Celsius.

## What are some common applications of energy requirement from water-steam?

The energy requirement from water-steam is a crucial aspect of many industrial processes, such as power generation, heating systems, and chemical reactions. It is also used in household appliances like kettles and steam irons.

## How can we reduce the energy requirement from water-steam?

There are several ways to decrease the energy requirement from water-steam, including using more efficient equipment and insulation, recycling waste heat, and implementing energy-saving practices such as reducing steam leaks and optimizing steam usage. Additionally, using renewable energy sources to heat water can significantly reduce the overall energy requirement.

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