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Water is being heated in a closed pan on top of a range while being stirred by a paddle wheel. During the process, 30 kJ of heat is transferred to the water, and 5 kJ of heat is lost to the surrounding air. The paddle-wheel work amounts to 5000 N m. Determine the final energy of the system if its initial energy is 10 kJ.

Respuesta :

Answer:

Final energy will be 40 KJ            

Explanation:

We have given that 30 KJ of the heat is transferred to the water

So [tex]Q_1=30kj[/tex]

And 5 KJ of heat is lost to the surrounding air  

So [tex]Q_2=-5kj[/tex]

So total heat [tex]Q=Q_1+Q_2=30+(-5)=25KJ[/tex]

The paddle-wheel work amounts to 5000 N -m

So work done W = -5 KJ ( as work is done by the paddle wheel )

We know that [tex]Q=W+\Delta U[/tex]

[tex]25=-5+\Delta U[/tex]

[tex]\Delta U=30KJ[/tex]

We have given initial kinetic energy = 10 KJ

So [tex]final -initia=30[/tex]

[tex]final-10=30[/tex]

So final  energy = 30+10 =40 KJ

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