A mixture of 1773 g of water and 227 g of ice is in an initial equilibrium state at 0.000°C. The mixture is then, in a reversible process, brought to a second equilibrium state where the water–ice ratio, by mass, is 1.00:1.00 at 0.000°C. (a) Calculate the entropy change of the system during this process. (The heat of fusion for water is 333 kJ/kg.) (b) The system is then returned to the initial equilibrium state in an irreversible process (say, by using a Bunsen burner). Calculate the entropy change of the system during this process, (c) Are your answers consistent with the second law of thermodynamics ?
A mixture of 1773 g of water and 227 g of ice is in an initial equilibrium state at 0.000°C. The mixture is then, in a reversible process, brought to a second equilibrium state where the water–ice ratio, by mass, is 1.00:1.00 at 0.000°C. (a) Calculate the entropy change of the system during this process. (The heat of fusion for water is 333 kJ/kg.) (b) The system is then returned to the initial equilibrium state in an irreversible process (say, by using a Bunsen burner). Calculate the entropy change of the system during this process, (c) Are your answers consistent with the second law of thermodynamics ?
A mixture of 1773 g of water and 227 g of ice is in an initial equilibrium state at 0.000°C. The mixture is then, in a reversible process, brought to a second equilibrium state where the water–ice ratio, by mass, is 1.00:1.00 at 0.000°C. (a) Calculate the entropy change of the system during this process. (The heat of fusion for water is 333 kJ/kg.) (b) The system is then returned to the initial equilibrium state in an irreversible process (say, by using a Bunsen burner). Calculate the entropy change of the system during this process, (c) Are your answers consistent with the second law of thermodynamics?
Science that deals with the amount of energy transferred from one equilibrium state to another equilibrium state.
Required information
An object is placed 20.0 cm from a converging lens with focal length 15.0 cm (see the figure, not drawn to scale). A
concave mirror with focal length 10.0 cm is located 76.5 cm to the right of the lens. Light goes through the lens, reflects
from the mirror, and passes through the lens again, forming a final image.
Converging
lens
Object
Concave
mirror
15.0 cm
-20.0 cm-
10.0 cm
d cm
d = 76.5.
What is the location of the final image?
cm to the left of the lens
!
Required information
A man requires reading glasses with +2.15-D refractive power to read a book held 40.0 cm away with a relaxed eye.
Assume the glasses are 1.90 cm from his eyes.
His uncorrected near point is 1.00 m. If one of the lenses is the one for distance vision, what should the refractive power of the other
lens (for close-up vision) in his bifocals be to give him clear vision from 25.0 cm to infinity?
2.98 D
!
Required information
Assume that the magnifier is held close to the eye. Use the standard near point of 25.0 cm to find the angular
magnification. An insect that is 4.10 mm long is placed 10.3 cm from a simple magnifier with a focal length of 13.0 cm.
What is the angular magnification?
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
The Second Law of Thermodynamics: Heat Flow, Entropy, and Microstates; Author: Professor Dave Explains;https://www.youtube.com/watch?v=MrwW4w2nAMc;License: Standard YouTube License, CC-BY