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.

Question

Nathalie Strickland

Physics

7 November, 12:06

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.

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Answers (1)

Know the Answer?

Bryan Berry · Answer 1

1.04 x 107 J.

Explanation:

We can use the following method to do the calculation

Total energy given to water to convert intosteam

dQ = m * l

dQ = 5.00 * 2.26 * 106

= 1.13 * 107 J

Work done at constantpressure dW = P * dV

Initialvolume V1 = 5.00kg / 958

= 5.22 * 10-3 m3

Finalvolume = 8.50 m3

=> dW = 1.01 * 105 * (8.50 - 5.22 * 10-3)

= 8.58 * 105 J

First law of thermodynamicsis dQ = ΔU + dW

Change in internalenergy ΔU = 1.13 * 107 - 8.58 * 105

= 1.04 x 107 J as our answer