Suppose 2 kg of water at 300 kPa has a specific volume of = 0 : 5m^3/kg. As a re

Question

Suppose 2 kg of water at 300 kPa has a specific volume of = 0 : 5m^3/kg. As a result of heating at constant pressure the temperature rises to 700 C. What is the initial temperature and initial internal energy? What is the change in the volume of the system and the change in the internal energy of the system? Suppose 2 kg of water at 300 kPa has a specific volume of = 0 : 5m^3/kg. As a result of heating at constant pressure the temperature rises to 700 C. What is the initial temperature and initial internal energy? What is the change in the volume of the system and the change in the internal energy of the system?

Explanation / Answer

Change in heat = Change in internal energy +work done
dQ = dU + P(dV)

hence

dU=580
dV=(810-580)/150000= 1.53 * 10^-3 m^3

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