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5.00 kg of liquid water is heated to 100.0 °C in a closed system. At this temperature, the density of liquid water is 958 kg/m3. The pressure is maintained at atmospheric pressure of 1.01 × 105 Pa. A moveable piston of negligible weight rests on the surface of the water. The water is then converted to steam by adding an additional amount of heat to the system. When all of the water is converted, the final volume of the steam is 8.50 m3. The latent heat of vaporization of water is 2.26 × 106 J/kg. How much heat is added to the system in the isothermal process of converting all of the water into steam?

Respuesta :

Answer:

[tex]1.04\times 10^7\ J.[/tex]

Explanation:

In the question given :

Pressure is constant

Therefore, Work done, [tex]W=P\times\Delta V[/tex]

Pressure, P=1.01 × 105 Pa.

Final volume, [tex]V_f=8.50\ m^3.[/tex]

Initial volume, [tex]V_i=\dfrac{Mass}{density}=\dfrac{5}{958}=5.22\times10^-3\ m^3.[/tex]

Therefore, W=8.58\times 10^{5}\ J.

Also, Heat Given, [tex]Q=m\times L=5\times 2.26\times 10^{6}\ J=1.13\times 10^7\ J.[/tex]

Also, according to First law of thermodynamics:

[tex]\Delta U=Q-W=(1.13\times 10^7)-(8.58\times 10^5)=1.04\times 10^7\ J.[/tex]

Hence, this is the required solution.

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