Question

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 1

Answer:
`3590\ kJ`

Step-by-step explanation:

Given

Paddle wheel work is
`W=-5090\ kJ\quad \text{work is done on the system}`

Heat transfer from the tank is
`Q=-1500\ kJ\quad \text{heat taken from the system}`

From the first law of thermodynamics

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

`\Rightarrow \Delta U=Q-W\\\Rightarrow \Delta U=-1500-(-5090)\\\Rightarrow \Delta U=3590\ kJ`

Therefore, the change in internal energy is
`3590\ kJ`