Calculate the change in internal energy of 2kg of water at 90 degree Celsius when it is changed to

Question

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

Answer 1

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

Answer 2

45.72mJ

Answer 3

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