A closed system of mass 20 kg undergoes a process in which there is a heat transfer of 1000 Q6: ki from the system to the surroundings. The work done on the system is 200 kl. If the initial 5 specific internal energy of the system is (250+R:) kl/kg, what is the final specific internal energy, in kj/kg? Neglect changes in kinetic and potential energy:

The final specific internal energy : 190 kJ/kg

The laws of thermodynamics 1 state that: energy can be changed but cannot be destroyed or created  

The equation is:  

Energy owned by the system is expressed as internal energy (U)  

This internal energy can change if it absorbs heat Q (U> 0), or releases heat (U <0). Or the internal energy can change if the system does work or accepts work (W)  

The sign rules for heat and work are set as follows:  

• The system receives heat, Q +  

• The system releases heat, Q -  

• The system does work, W -  

• the system accepts work, W +  

A closed system of mass 20 kg⇒m=20 kg

Heat transfer of 1000 kJ from the system to the surroundings⇒Q=-1000 kJ

The work done on the system is 200 kJ⇒W=+200 kJ

The initial specific internal energy of the system is 250 kJ /kg⇒U₁ = 250 kj/kg

Neglect changes in kinetic and potential energy⇒ΔKE+ΔPE=0, so

Q-W = ΔU

Input in equation

answer:

an example of solubility is salt or sugar dissolving in water.

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