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A tank containing a fluid is stirred by a paddle wheel. The work input to the paddle wheel is 5090 kJ. The heat transfer from the tank is 1500 kJ. Consider the tank and the fluid inside a control surface and determine the change in internal energy, in kJ, of this control mass.

Answer :

Answer: [tex]3590\ kJ[/tex]

Explanation:

Given

Paddle wheel work is [tex]W=-5090\ kJ\quad \text{work is done on the system}[/tex]

Heat transfer from the tank is [tex]Q=-1500\ kJ\quad \text{heat taken from the system}[/tex]

From the first law of thermodynamics

Change in the internal energy of the system is equal to the difference of heat and work .

[tex]\Rightarrow \Delta U=Q-W\\\Rightarrow \Delta U=-1500-(-5090)\\\Rightarrow \Delta U=3590\ kJ[/tex]

Therefore, the change in internal energy is [tex]3590\ kJ[/tex]

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