Question about the 2nd law of thermodynamics (heat loss)

AI Thread Summary
In a closed system with an ideal gas and a movable piston, the first law of thermodynamics states that the heat added (dQ) equals the change in internal energy (dU) plus the work done (dW). The second law of thermodynamics indicates that while mechanical energy can be fully converted to heat, the reverse is not entirely possible. When heat is added to the cylinder at constant temperature, some energy is lost due to inefficiencies in the conversion process. Although heat energy can be converted to work, this cannot be achieved in a cyclic process without some energy loss. Thus, maintaining constant temperature while adding heat results in energy dissipation, highlighting the limitations of energy conversion in thermodynamic systems.
Shovon00000
Messages
3
Reaction score
0
Thread moved from the technical forums to the schoolwork forums
Assume that a closed system of cylindar filled with ideal gas consists of a movable piston.We know from the 1st law dQ=dU +dW.
According to the 2nd law mechanical energy can be totally converted into heat energy but heat energy cannot be converted completely into mechanical energy.The question is"If we give heat to the cylindar and keep the temperature constant how will some of the heat energy be lost ?(here dU=0,and we know dQ=dU+dW)
 
Physics news on Phys.org
You can completely convert heat energy to work. You just can't do it by operating the system in a cycle.
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top