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

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The calculation for thermal energy in mixing water and steam is confirmed as accurate, yielding a total energy of 7.12 million joules. The formula used combines the heat transfer from warm water and the latent heat of steam. The discussion clarifies that only 2 kg of water is converted to steam, as specified in the problem statement. Attention to significant digits and proper prefix usage is emphasized for precision. Overall, the calculations align with the principles of thermodynamics.
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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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If have close pipe system with water inside pressurized at P1= 200 000Pa absolute, density 1000kg/m3, wider pipe diameter=2cm, contraction pipe diameter=1.49cm, that is contraction area ratio A1/A2=1.8 a) If water is stationary(pump OFF) and if I drill a hole anywhere at pipe, water will leak out, because pressure(200kPa) inside is higher than atmospheric pressure (101 325Pa). b)If I turn on pump and water start flowing with with v1=10m/s in A1 wider section, from Bernoulli equation I...

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