Is the Calculation of Thermal Energy in Mixing Water and Steam Accurate?

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SUMMARY

The calculation of thermal energy in mixing water and steam is accurately represented by the equation Q=Qwarm water+Qsteam, where Q is the total heat energy. The specific heat capacity of water (c) is 4200 J/kg°C, and the latent heat of vaporization (Lv) for steam is 2.3 x 10^6 J/kg. In the provided example, the total energy calculated is Q=7.12 x 10^6 J, based on 8 kg of water cooling from 100°C to 25°C and 2 kg of steam condensing. The answer is confirmed as correct, emphasizing the importance of significant digits and proper unit prefixes.

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Homework Statement
How much energy is required to change one quarter of 8.0 kg of water at 25 degrees celsuis into steam(100 degrees celcuis)
Relevant Equations
Q=mct
Q=mLv
Q=Qwarm water+Qsteam
Q=mct+mLv
Q=(8kg)(4200J)(100C-25C)+(2kg)(2.3*10 to the power of 6 J/Kg)
Q=7.12*10 to the power of 6 J
Is this the correct answer?
 
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Why only 2kg water turning to steam?
 
Because that's what the question states
 
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Ah great, I missed the one quarter.

Answer is correct. Keep in mind correct significant digits and usage of prefix
 
Last edited:
Thank you!
 
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