Recent content by beetlejuice

Okay, I see how those things add up. Thank you! What would I do to calculate the number of marbles needed to calculate that change? For the closed system, I used the first law. Does the same method still apply?

1. Homework Statement We want to heat up 200 liters of water in a bathtub from 15 ◦C to 30 ◦C. The temperature is increased by adding marble stones to the water. The initial temperature of the marble stones is 773, 15 K. Assume that the marble is rigid, with the heat capacity cm = 0.88 J ·K−1...

Ah okay, so W= -309.98kJ (?)

Oh, so k=n. Now I understand. W = -0.02154J

Is k=PV? As for finding volume values, I can use pV=mRT where (0.7224)(V1)=(15)(0.285)(300) That gives us V1= 0.01775 Then I can apply p1(V1^n)=p2(V2^n) where n=1.41 to solve for V2. V2= 0.01409 Then I guess the rest is plug and chug? (Assuming I'm on the right path here...)

I'm not sure I do. I have the equation that W= -p*dV, but I am not sure if that's correct

What I mean is the force from the net pressure is in the upwards direction (acting against gravity). I have F= -19.62N (downwards being negative, force from the weight of the piston). Total P = F/A = -19.62/(π*(0.015^2)) = -0.2776 bar External pressure is 1 bar and thus I now believe that...

gravity down (naturally), piston acts upwards but is stationed at the bottom, the internal pressure would pull the piston upwards, we are assuming the system is at rest and no mechanical forces are acting on the piston

It would be isentropic! Also, I have recently found out that my original answer to the calculation of the internal pressure was incorrect. Is there any chance you could also help me with that part? Originally, I put p=F/A= (5kg*9.81ms^2)/(pi*0.015^2)= 27.8kPa. But, now I am not confident in the...

[Mentor's note: moved from another forum, so homework template missing.] An upright and ideally heat-insulated cylinder with a diameter of 30 mm is tightly sealed at the bottom by a piston of the mass mk = 2 kg. The cylinder contains 15 g of air. The ambient pressure is pu = 1 bar, the initial...