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phy

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Calculate the change in internal energy of 2kg of water at 90 degree Celsius when it is changed to
3.30m3
of steam at
100oC
. The whole process occurs at atmospheric pressure. The latent heat of vaporization of water is
2.26×106J/kg

  • phy -

    ΔQ=ΔU+W
    ΔU = ΔQ – W =
    =c•m•ΔT + r•m – p•ΔV=
    = c•m•ΔT + r•m – p{V- (m/ρ)} =
    =4190•2•10 +2.26•10⁶•2 – 101325(3.3 – 2/1000) = 4515630 J

  • phy -

    45.72mJ

  • phy -

    Answer is du= mc(teta) + ml - pdv. Solving it gives you 4269830.2J
    Find me @word_geniusoche

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