# 15.3 Introduction to the second law of thermodynamics: heat engines  (Page 5/10)

 Page 5 / 10

The net work done by a cyclical process is the area inside the closed path on a $\text{PV}$ diagram, such as that inside path ABCDA in [link] . Note that in every imaginable cyclical process, it is absolutely necessary for heat transfer from the system to occur in order to get a net work output. In the Otto cycle, heat transfer occurs along path DA. If no heat transfer occurs, then the return path is the same, and the net work output is zero. The lower the temperature on the path AB, the less work has to be done to compress the gas. The area inside the closed path is then greater, and so the engine does more work and is thus more efficient. Similarly, the higher the temperature along path CD, the more work output there is. (See [link] .) So efficiency is related to the temperatures of the hot and cold reservoirs. In the next section, we shall see what the absolute limit to the efficiency of a heat engine is, and how it is related to temperature.

## Section summary

• The two expressions of the second law of thermodynamics are: (i) Heat transfer occurs spontaneously from higher- to lower-temperature bodies but never spontaneously in the reverse direction; and (ii) It is impossible in any system for heat transfer from a reservoir to completely convert to work in a cyclical process in which the system returns to its initial state.
• Irreversible processes depend on path and do not return to their original state. Cyclical processes are processes that return to their original state at the end of every cycle.
• In a cyclical process, such as a heat engine, the net work done by the system equals the net heat transfer into the system, or $W={Q}_{\text{h}}–{Q}_{\text{c}}\phantom{\rule{0.25em}{0ex}}$ , where ${Q}_{\text{h}}$ is the heat transfer from the hot object (hot reservoir), and ${Q}_{\text{c}}$ is the heat transfer into the cold object (cold reservoir).
• Efficiency can be expressed as $\text{Eff}=\frac{W}{{Q}_{\text{h}}}$ , the ratio of work output divided by the amount of energy input.
• The four-stroke gasoline engine is often explained in terms of the Otto cycle, which is a repeating sequence of processes that convert heat into work.

#### Questions & Answers

what is physics
a15kg powerexerted by the foresafter 3second
what is displacement
movement in a direction
Jason
hello
Hosea
Explain why magnetic damping might not be effective on an object made of several thin conducting layers separated by insulation? can someone please explain this i need it for my final exam
Hi
saeid
hi
Yimam
What is thê principle behind movement of thê taps control
while
Hosea
what is atomic mass
this is the mass of an atom of an element in ratio with the mass of carbon-atom
Chukwuka
show me how to get the accuracies of the values of the resistors for the two circuits i.e for series and parallel sides
Explain why it is difficult to have an ideal machine in real life situations.
tell me
Promise
what's the s . i unit for couple?
Promise
its s.i unit is Nm
Covenant
Force×perpendicular distance N×m=Nm
Oluwakayode
İt iş diffucult to have idêal machine because of FRİCTİON definitely reduce thê efficiency
Oluwakayode
if the classica theory of specific heat is valid,what would be the thermal energy of one kmol of copper at the debye temperature (for copper is 340k)
can i get all formulas of physics
yes
haider
what affects fluid
pressure
Oluwakayode
Dimension for force MLT-2
what is the dimensions of Force?
how do you calculate the 5% uncertainty of 4cm?
4cm/100×5= 0.2cm
haider
how do you calculate the 5% absolute uncertainty of a 200g mass?
= 200g±(5%)10g
haider
use the 10g as the uncertainty?
melia
which topic u discussing about?
haider
topic of question?
haider
the relationship between the applied force and the deflection
melia
sorry wrong question i meant the 5% uncertainty of 4cm?
melia
its 0.2 cm or 2mm
haider
thank you
melia
Hello group...
Chioma
hi
haider
well hello there
sean
hi
Noks
hii
Chibueze
10g
Olokuntoye
0.2m
Olokuntoye
hi guys
thomas