Recent content by songoku

The change from the condition where the object is water at 100oC to the condition where the object is vaopur at 100oC Since internal energy is the sum of kinetic and potential energy and for both cases the change in KE and PE are the same so the change in internal energy is also the same

Case (1) = pressure is equal to atmospheric pressure Case (2) = pressure is higher than atmospheric pressure Pressure higher than atmospheric pressure will increase the boiling point of water. Water molecules in case (2) would have higher KE than in case (1) because they are at higher...

I do not learn Clapeyron Clausius equation @kuruman what is the mistake in post #19?

I am not sure what you mean. ΔU = Q + W where W is work done on the system Q = m L where L is the specific latent heat of vaporization and is the one we want to find out ΔU = ΔKE + ΔPE = 1/2 m (v22 - v12) + ΔPE Sorry I don't know the formula of ΔPE for this case. I just know PE in this case...

Increasing the pressure for same ΔV will increase the work done by the system so the work done on the system will decrease (will be more negative) ΔU = change in internal energy of water ΔKE = change in kinetic energy of water molecules ΔPE = change in potential energy of water molecules

Because work done by the system = p . ΔV so higher p means higher W I wrote the formula as ΔU = Q + W so shouldn't the W there is work done on the system? My idea is to find comparison of Q and Q' through the comparison of ΔU and ΔU' and what I can think of is using ΔU = ΔKE + ΔPE ΔU' = ΔKE'...

I think so. By taking ΔV to the same, we can say that the work done by the system will be higher. Let me try: For same ΔV, work done by system is higher so the work done on system will be more negative. When the pressure is higher, the boiling point of the water increases so the particles will...

ΔU = Q + W Q = ΔU - W I am not sure how to analyze the changes in ΔU and W. The value of p increases, but how do we know about ΔV? For ΔU, the KE is constant and the PE increases since the molecules' separation is more but how do we compare the increase to the case of normal atmospheric...

Thank you very much for the explanation Rap and Gavran

I have found part (a): current on the left is 7/12 A from A to B current on the middle is 5/36 A from F to C current on the right is 13/18 A from D to E For part (b), what is the meaning of VBF? Is it VB - VF or VF - VB? My attempt is: VBF = I1 x R1 - ε1 = 7/12 x 4 - 1.5 = 5/6 V Is it...

Sorry for late reply Thank you very much for the help and explanation berkeman, kuruman, Steve4Physics, Gavran

Sorry I got the question in form of printed paper so I don't have the link. I think the answer is (d) but I am not sure how the distribution (shape) of magnetic field inside the steel sheet would be. Is there any possibility some part of the magnetic field is redistributed by the steel sheet...

The answer key is (a) but I don't understand why. Option (a), (b), (c) are all non - magnetic material and (d) is magnetic material. Why (a) can stop the attraction while others can't? Thanks

Thank you very much for all the help and explanation haruspex, wrobel, Herman Trivilino, 256bits, Steve4Physics, jbriggs444

avoid making the assumption and use all the information provided. For this question, consideration about breaking stress was not in my head until I read your reply