(III) A general theorem states that the amount of energy that becomes unavailable to do useful work in any process is equal to T L ∆ S , where T L is the lowest temperature available and ∆ S is the total change in entropy during the process. Show that this is valid in the specific cases of ( a ) a falling rock that comes to rest when it hits the ground; ( b ) the free adiabatic expansion of an ideal gas; and ( c ) the conduction of heat, Q , from a high-temperature ( T H ) reservoir to a low-temperature ( T L ) reservoir. [ Hint : In part ( c ) compare to a Carnot engine.]
(III) A general theorem states that the amount of energy that becomes unavailable to do useful work in any process is equal to T L ∆ S , where T L is the lowest temperature available and ∆ S is the total change in entropy during the process. Show that this is valid in the specific cases of ( a ) a falling rock that comes to rest when it hits the ground; ( b ) the free adiabatic expansion of an ideal gas; and ( c ) the conduction of heat, Q , from a high-temperature ( T H ) reservoir to a low-temperature ( T L ) reservoir. [ Hint : In part ( c ) compare to a Carnot engine.]
(III) A general theorem states that the amount of energy that becomes unavailable to do useful work in any process is equal to TL ∆S, where TL is the lowest temperature available and ∆S is the total change in entropy during the process. Show that this is valid in the specific cases of (a) a falling rock that comes to rest when it hits the ground; (b) the free adiabatic expansion of an ideal gas; and (c) the conduction of heat, Q, from a high-temperature (TH) reservoir to a low-temperature (TL) reservoir. [Hint: In part (c) compare to a Carnot engine.]
L₁
D₁
L₂
D2
Aluminum has a resistivity of p = 2.65 × 10 8 2. m. An aluminum wire is L = 2.00 m long and has a
circular cross section that is not constant. The diameter of the wire is D₁ = 0.17 mm for a length of
L₁ = 0.500 m and a diameter of D2 = 0.24 mm for the rest of the length.
a) What is the resistance of this wire?
R =
Hint
A potential difference of AV = 1.40 V is applied across the wire.
b) What is the magnitude of the current density in the thin part of the wire?
Hint
J1
=
c) What is the magnitude of the current density in the thick part of the wire?
J₂ =
d) What is the magnitude of the electric field in the thin part of the wire?
E1
=
Hint
e) What is the magnitude of the electric field in the thick part of the wire?
E2
=
please help
A cheetah spots a gazelle in the distance and begins to sprint from rest, accelerating uniformly at a rate of 8.00 m/s^2 for 5 seconds. After 5 seconds, the cheetah sees that the gazelle has escaped to safety, so it begins to decelerate uniformly at 6.00 m/s^2 until it comes to a stop.
Chapter 20 Solutions
Physics for Scientists and Engineers with Modern Physics
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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