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

`\Delta U = Q-W = -150 KJ -75KJ= -225 KJ`

`\Delta u = (\Delta U)/(m)=(-225 KJ)/(3Kg)= -75 (KJ)/(Kg)`

Step-by-step explanation:

For this case we know the mass of the system
`m = 3kg`

For this case the work done by the system is +75 KJ and is positive since we assume that the work is done on the sytem by the surroundings

And the heat for this case would be negative since is transferred from the system to the sorroundings Q = -150 KJ

From the problem we know that the can neglect the potential and the kinetic energy so then we have that:

`\Delta PE= \Delta KE = 0`

Using the first law of thermodynamics we know this:

`E_(in)- E_(out)= \Delta E`

Where E means energy, and if we replace we have:

`Q-W = \Delta KE + \Delta PE + \Delta U`

So then we have just this formula:

`Q-W = \Delta U`

`\Delta U = Q-W = -150 KJ -75KJ= -225 KJ`

And then since we need the specific internal energy we can do this:

`\Delta u = (\Delta U)/(m)=(-225 KJ)/(3Kg)= -75 (KJ)/(Kg)`