Question

A closed system of mass 3 kg undergoes a process in which there is a heat transfer of 150 kJ from the system to the surroundings. The work done on the system is 75 kJ. If the initial specific internal energy of the system is 450 kJ/kg, what is the final specific internal energy, in kJ/kg? Neglect changes in kinetic and potential energy.

Answer 1

`u_(2) = 425\,(kJ)/(kg)`

Step-by-step explanation:

The Energy Balance is modelled after the First Law of Thermodynamics for a closed system:

`-Q_(out) + W_(in) = m\cdot (u_(2)-u_(1))`

The final specfic internal energy is cleared inside the formula:

`u_(2)=u_(1)+(W_(in)-Q_(out))/(m)`

`u_(2) = 450\,(kJ)/(kg) + (75\,kJ-150\,kJ)/(3\,kg)`

`u_(2) = 425\,(kJ)/(kg)`