@c -*-texinfo-*-
@c This is part of the FisicaLab User Manual.
@c Copyright (C)  2023 German A. Arias.
@c See the file copying.texinfo for copying conditions.

@node Examples thermodynamics, GFDL, Thermodynamics, Top
@chapter Examples thermodynamics

@cindex Examples thermodynamics


@menu
* Example 1 (thermo)::             
* Example 2 (thermo)::             
* Example 3 (thermo)::             
* Example 4 (thermo)::             
* Example 5 (thermo)::             
@end menu


@node Example 1 (thermo), Example 2 (thermo), Examples thermodynamics, Examples thermodynamics
@section Example 1

@cindex Example 1 (thermo)

700 cal are given to a thermodynamic process, while the system made a
work of 200 J. How many change the internal energy?

@sp 2

@strong{Solution with FisicaLab}

The problem statement don't specify the type of thermodynamics
process, so we can assume it is an isobaric process witn an arbitrary
mass, pressure and initial temperature. Then select the Thermodynamics
group and, inside this, the Thermodynamics module. Erase the content
of the chalkboard. And add one element Isobaric process, one element
Applied heat and one element Work. As show the image below:

@center @image{therm-e1, 3cm}

To the isobaric process, we set 1 kg to mass, 1 pascal to pressure and
1 m^3 to initial volumen. Then:

@sp 1

@table @strong

@item Name
0

@item m
1

@item dU
U

@item P
1

@item Vi
1

@item Vf
Vf

@end table

@sp 1

To the element Applied heat:

@sp 1

@table @strong

@item Q
700 @@ cal

@end table

@sp 1

And to the element Work:

@sp 1

@table @strong

@item W
200

@end table

@sp 1

Now click in the icon Solve to get the answer:

@example

U = 2312.080  J ;
Vf  =  201.000  m3;
Status = success.

@end example

@page


@node Example 2 (thermo), Example 3 (thermo), Example 1 (thermo), Examples thermodynamics
@section Example 2

@cindex Example 2 (thermo)

A work of 200 J is done over a system through an adiabatic
process. Calculate the change at internal energy in calories.

@sp 2

@strong{Solution with FisicaLab}

Select the Thermodynamics group and, inside this, the Thermodynamics
module. Erase the content of the chalkboard. And add one element
Adiabatic process and one element Work. As show the image below:

@center @image{therm-e2, 3cm}

To the element Adiabatic process, with an arbitrary mass of 1 kg, we
have:

@sp 1
 
@table @strong

@item Name
0

@item m
1

@item dU
U @@ cal

@end table

@sp 1

Now to the element Work:

@sp 1

@table @strong

@item W
-200

@end table

@sp 1

Now click in the icon Solve to get the answer:

@example

U  =  47.769  cal ;
Status = success.

@end example

@page


@node Example 3 (thermo), Example 4 (thermo), Example 2 (thermo), Examples thermodynamics
@section Example 3

@cindex Example 3 (thermo)

At a constant pressure of 1 atm, 1.5 g of water (1.5 cm^3) at 100 °C
is vaporized to a final volume of 2000 cm^3. Calculate the work done
by the system and its change of internal energy.

@sp 2

@strong{Solution with FisicaLab}

Select the Thermodynamics group and, inside this, the Thermodynamics
module. Erase the content of the chalkboard. And add one element
Isobaric process, one element Applied heat, one element Work and one
element Vaporization heat. As show the image below:

@center @image{therm-e3, 3cm}

To the element Isobaric process, we have:

@sp 1

@table @strong

@item Name
proc

@item m
1.5 @@ g

@item dU
U

@item P
1 @@ atm

@item Vi
1.5 @@ cm3

@item Vf
2000 @@ cm3

@end table

@sp 1

To the element Applied heat:

@sp 1

@table @strong

@item Q
Q

@end table

@sp 1

To the element Work:

@sp 1

@table @strong

@item W
W

@end table

@sp 1

And to the element vaporized heat:

@sp 1

@table @strong

@item Process:
proc

@item cv:
540 @@ cal/g

@end table

@sp 1

Now click in the icon Solve to get the answer:

@example

W  =  202.498  J ;
U  =  3188.810  J ;
Q  =  3391.308  J ;
Status = success 

@end example

@page


@node Example 4 (thermo), Example 5 (thermo), Example 3 (thermo), Examples thermodynamics
@section Example 4

@cindex Example 4 (thermo)

A steam machine works beetwen an high temperature of 230 °C and a
lower temperature of 100 °C. Calculate its ideal efficiency.

@sp 2

@strong{Solution with FisicaLab}

Select the Thermodynamics group and, inside this, the Thermodynamics
module. Erase the content of the chalkboard and add one element Heat
engine, one element Applied heat, one element Extracted heat and one
element Work. As show the image below:

@center @image{therm-e4, 3cm}

To the element Heat machine, we have:

@sp 1

@table @strong

@item TH
230 @@ C

@item TL
100 @@ C

@item e
e

@end table

@sp 1

To the element Applied heat set any arbitrary value (positive):

@sp 1

@table @strong

@item Q
1000

@end table

@sp 1

To the element Extracted heat:

@sp 1

@table @strong

@item Q
Ex

@end table

@sp 1

And to the element Work:

@sp 1

@table @strong

@item W
W

@end table

@sp 1

Now click in the icon Solve to get the answer:

@example

W  =  258.372  J ;
e  =  25.837  % ;
Ex  =  741.628  J ;
Status = success.

@end example

@page


@node Example 5 (thermo), GFDL, Example 4 (thermo), Examples thermodynamics
@section Example 5

@cindex Example 5 (thermo)

At a refrigerator with a coefficient of performance of 2.18 and 680 J of
internal work, calculate the amount of heat (in Joules) extracted at the
could reservoir and the amount of heat transferred to the hot reservoir.

@sp 2

@strong{Solution with FisicaLab}

Select the Thermodynamics group and, inside this, the Thermodynamics
module. Erase the content of the chalkboard. And add one element
Refrigerator, one element Applied heat, one element Heat extracted and
one element Work. As show the image below:

@center @image{therm-e5, 3cm}

To the element Refrigerator set an arbitrary low temperature, we have:

@sp 1

@table @strong

@item TH
H

@item TL
100 @@ C

@item cop
2.18

@end table

@sp 1

To the element Applied heat:

@sp 1

@table @strong

@item Q
Q

@end table

@sp 1

To the element Extracted heat:

@sp 1

@table @strong

@item Q
Ex

@end table

@sp 1

And to the element Work:

@sp 1

@table @strong

@item W
680

@end table

@sp 1

Then click in the icon Solve to get the answer:

@example

Ex  =  1482.400  J ;
H  =  271.170  C ;
Q  =  2162.400  J ;
Status = success.

@end example
