Energy required to make 2.0 kg of steam

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SUMMARY

To convert 2.0 kg of water at 25 ˚C into steam, a total energy of 7.1 x 106 J must be added. This calculation involves warming the water to 100 ˚C and then boiling it, using the specific heat capacity of water (4200 J/(kg·°C)) and the latent heat of vaporization (2.3 x 106 J/kg). The total energy is derived from the equation Q_total = Q_warm water + Q_boil water, resulting in 2,520,000 J for warming and 4,600,000 J for boiling.

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Homework Statement


A container holds 8.0 kg of water at 25 ˚C. How much energy must be added to the water to create 2.0 kg of steam?

Homework Equations


To be honest, not sure. SO MANY equations were given, and I don't know what is a variation of what.

The Attempt at a Solution


Q_total = Q_warm water + Q_boil water
=mc_wt+m_sL_v
= (8.0 kg)(4200 J/(kg* degrees C))(100 degrees C - 25 degrees C)+(2.0 kg)(2.3x10^6)
=2520000 J + 4600000 J
=7120000 J
=7.1 x 10^6 J

Therefore 7.1 x 10^6 J of heat energy have to be added to the water to create 2.0 kg of steam.

Does that make any sense?
 
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LonelyElectron said:
Does that make any sense?
Yes, it makes a lot of sense.
 
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kuruman said:
Yes, it makes a lot of sense.

Okay perfect! So its correct? You guys are the best. I hope you know how much your help is appreciated!
 
LonelyElectron said:
So its correct?
Yes, it's correct.
 
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