An ideal refrigerator or ideal heat pump is equivalent to a Carnot engine running in reverse. That is, energy | Q c | is taken in from a cold reservoir and energy | Q h | is rejected to a hot reservoir. (a) Show that the work that must he supplied to run the refrigerator or heat pump is W = T h − T c T c | Q c | (b) Show that the coefficient of performance (COP) of the ideal refrigerator is COP = T c T h − T c
An ideal refrigerator or ideal heat pump is equivalent to a Carnot engine running in reverse. That is, energy | Q c | is taken in from a cold reservoir and energy | Q h | is rejected to a hot reservoir. (a) Show that the work that must he supplied to run the refrigerator or heat pump is W = T h − T c T c | Q c | (b) Show that the coefficient of performance (COP) of the ideal refrigerator is COP = T c T h − T c
Solution Summary: The author explains how the work supplied to run the refrigerator is cWleft|Q_hright
An ideal refrigerator or ideal heat pump is equivalent to a Carnot engine running in reverse. That is, energy |Qc| is taken in from a cold reservoir and energy |Qh| is rejected to a hot reservoir. (a) Show that the work that must he supplied to run the refrigerator or heat pump is
W
=
T
h
−
T
c
T
c
|
Q
c
|
(b) Show that the coefficient of performance (COP) of the ideal refrigerator is
An ideal refrigerator or ideal heat pump is equivalent to a
Carnot engine running in reverse. That is, energy |Q] is
taken in from a cold reservoir and energy |Q,| is rejected
to a hot reservoir. (a) Show that the work that must be sup-
plied to run the refrigerator or heat pump is
Т, — т.
-
W =
T.
(b) Show that the coefficient of performance (COP) of the
ideal refrigerator is
T.
СОР
T- T.
It has been proposed to use the thermal gradient of the ocean to drive a heat engine. Suppose that at a certain location the water temperature is 22° C at the ocean surface and 40 C at the ocean floor.
(a) What is the maximum possible efficiency of an engine operating between these two temperatures? (b) If the engine is to produce 1 GW of electrical power, what minimum volume of water must be processed (to suck out the heat) in every second?
A power plant has been proposed that would make use of the temperature gradient in the ocean. The system is to operate between 20.0°C (surface water temperature) and 5.00°C (water temperature at a depth of about 1 km). (a) What is the maximum efficiency of such a system? (b) If the useful power output of the plant is 75.0 MW, how much energy is absorbed per hour? (c) In view of your answer to part (a), do you think such a system is worthwhile (considering that there is no charge for fuel)?
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