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A cylinder with a piston contains 0.2 \mathrm{~kg} of water at 100^{\circ} \mathrm{C}. What is the change in internal energy of the water when it is converted to steam at 100^{\circ} \mathrm{C} at a constant pressure of 1 atm? The density of water is \rho_o=10^3 \mathrm{~kg} / \mathrm{m}^3 and that of steam is \mathrm{\rho_s=0.6 \mathrm{~kg} / \mathrm{m}^3}. The latent heat of vaporization of water is \mathrm{L_v=2.26 \times 10^6 \mathrm{~J} / \mathrm{kg}.}
 

Option: 1

218.4 \mathrm{~kJ}


 


Option: 2

318.4 \mathrm{~kJ}
 


Option: 3

418.4 \mathrm{~kJ}
 


Option: 4

518.4 \mathrm{~kJ}


Answers (1)

best_answer

The heat transfer to the water is

\mathrm{Q=m L_v=(0.2 \mathrm{~kg})\left(2.26 \times 10^6 \mathrm{~J} / \mathrm{kg}\right)=4.52 \times 10^5 \mathrm{~J} }

The work done by the water when it expands against the piston at constant pressure is

\mathrm{ W=P\left(V_s-V_w\right)=P\left(\frac{m}{\rho_s}-\frac{m}{\rho_w}\right) }

\mathrm{ =\left(1.01 \times 10^5 \mathrm{~N} / \mathrm{m}^2\right)\left(\frac{0.2 \mathrm{~kg}}{0.6 \mathrm{~kg} / \mathrm{m}^3}-\frac{0.2 \mathrm{~kg}}{1000 \mathrm{~kg} / \mathrm{m}^3}\right) }

\mathrm{ =3.36 \times 10^4 \mathrm{~J} }

The change in internal energy is

\mathrm{ \Delta U=Q-W=452 \mathrm{~kJ}-33.6 \mathrm{~kJ}=418.4 \mathrm{~kJ} }
 

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Gunjita

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