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5.00 kg of liquid water at 100 kPa is heated from 20 C to 80 C...

5.00 kg of liquid water at 100 kPa is heated from 20 C to 80 C on the stove. Neglecting evaporation of the water, determine the entropy change of the water.

Assume cv= 4.20 kJ/(kg-K)

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Answer #1

Given - m=5kg Ti = 20 °C = 29315 T2 = 80°C = 353 K ev=4.2 kJ / kg K Entropy change faszinic, ln To je 12-si =584 12 ln ( 353

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