- #1
maitake91
- 9
- 4
- Homework Statement
- Heat is transferred through the wall of a vertical cylinder to air trapped inside it by a frictionless piston of mass 10kg and radius 0.05m. When 300J of heat is transferred to the air the piston rises by 0.2m against the atmospheric pressure of 100kPa. Determine the change in internal energy of the trapped air when 300J is transferred and the piston rises 0.2m.
- Relevant Equations
- (Qin - Qout)+(Win - Wout)+(Emassin - Emassout) = ∆U+∆KE+∆PE
After crossing out all the variables which I think equals 0 in the equation, I was left with:
∆PE + ∆U = Qin
mg (0.2) + ∆U = 300
10*9.81*0.2 - 300 = - ∆U
= -280J
This was the answer I derived. However, the correct answer was supposed to be 123.3J. Please can someone explain to me how to get the correct answer and where I did wrong? This may not be relevant but the pressure of the trapped air was worked out in earlier questions, which equals to 11240Pa.
∆PE + ∆U = Qin
mg (0.2) + ∆U = 300
10*9.81*0.2 - 300 = - ∆U
= -280J
This was the answer I derived. However, the correct answer was supposed to be 123.3J. Please can someone explain to me how to get the correct answer and where I did wrong? This may not be relevant but the pressure of the trapped air was worked out in earlier questions, which equals to 11240Pa.