5.00 kg of liquid water is heated to 100.0 °C in a closed system. At this temper

Question

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 x 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 x 106 J/kg. Calculate how much work is done and the change in the internal energy during this isothermal process.

I have no idea how to solve this! Please help and show formulas that are necessary in this problem!

Explanation / Answer

so this is a constant pressure problem. so W=P dV, where P =1.01E5 the final volume is 8.5 m^3 the initial volume we can find from density=M/V V=M/density=5/958=5.2E-3 m^3 so W=1.01E5*(8.5-5.2E-3)=8.58E5 J we also know Q=mL=5*2.26E6 J=1.13E7 J then we can use that dU=Q-W=1.13E7-8.58E5=1.04E7 J

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