# chapter 2:Zeroth Law of Thermodynamics¶

## example 2.1;pg no:46¶

In [1]:
#cal of temperature of body of human
#intiation of all variables
# Chapter 2
print"Example 2.1, Page:46  \n \n"
print("Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 1")
Tf=98.6;#temperature of body in farenheit
Tc=(Tf-32)/1.8
print("degree celcius and farenheit are related as follows")
print("Tc=(Tf-32)/1.8")
print("so temperature of body in degree celcius"),round(Tc,2)

Example 2.1, Page:46

Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 1
degree celcius and farenheit are related as follows
Tc=(Tf-32)/1.8
so temperature of body in degree celcius 37.0


## example 2.2;pg no:47¶

In [2]:
#cal of celcius temperature
#intiation of all variables
# Chapter 2
import math
print"Example 2.2, Page:47  \n \n"
print("Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 2")
t1=0;#ice point temperature in degree celcius
p1=3;#thermometric property for ice point
t2=100;#steam point temperature in degree celcius
p2=8;#thermometric property for steam point
p3=6.5;#thermometric property for any temperature
print("using thermometric relation")
print("t=a*log(p)+(b/2)")
print("for ice point,b/a=")
b=2*math.log(p1)
print("so b=2.1972*a")
print("for steam point")
a=t2/(math.log(p2)-(2.1972/2))
print("a="),round(a,2)
b=2.1972*a
print("and b="),round(b,2)
t=a*math.log(p3)+(b/2)
print("thus, t=in degree celcius")
print("so for thermodynamic property of 6.5,t=302.83 degree celcius="),round(t,2)

Example 2.2, Page:47

Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 2
using thermometric relation
t=a*log(p)+(b/2)
for ice point,b/a=
so b=2.1972*a
for steam point
a= 101.95
and b= 224.01
thus, t=in degree celcius
so for thermodynamic property of 6.5,t=302.83 degree celcius= 302.84


## example 2.3;page no:47¶

In [3]:
#cal of temperature shown by this thermometer
#intiation of all variables
# Chapter 2
print"Example 2.3, Page:47  \n \n"
print("Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 3")
print("emf equation")
print("E=(0.003*t)-((5*10^-7)*t^2))+(0.5*10^-3)")
print("using emf equation at ice point,E_0 in volts")
t=0.;#ice point temperature in degree celcius
E_0=(0.003*t)-((5.*10**-7)*t**2)+(0.5*10**-3)
print("E_0="),round(E_0,2)
print("using emf equation at steam point,E_100 in volts")
t=100.;#steam point temperature in degree celcius
E_100=(0.003*t)-((5.*10**-7)*t**2)+(0.5*10**-3)
print("E_100="),round(E_100,2)
print("now emf at 30 degree celcius using emf equation(E_30)in volts")
t=30.;#temperature of substance in degree celcius
E_30=(0.003*t)-((5.*10**-7)*t**2)+(0.5*10**-3)
T_100=100.;#steam point temperature in degree celcius
T_0=0.;#ice point temperature in degree celcius
T=((E_30-E_0)/(E_100-E_0))*(T_100-T_0)
print("now the temperature(T) shown by this thermometer")
print("T=in degree celcius"),round(T,2)
print("NOTE=>In this question,values of emf at 100 and 30 degree celcius is calculated wrong in book so it is corrected above so the answers may vary.")

Example 2.3, Page:47

Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 3
emf equation
E=(0.003*t)-((5*10^-7)*t^2))+(0.5*10^-3)
using emf equation at ice point,E_0 in volts
E_0= 0.0
using emf equation at steam point,E_100 in volts
E_100= 0.3
now emf at 30 degree celcius using emf equation(E_30)in volts
now the temperature(T) shown by this thermometer
T=in degree celcius 30.36
NOTE=>In this question,values of emf at 100 and 30 degree celcius is calculated wrong in book so it is corrected above so the answers may vary.


## example 2.4;pg no:48¶

In [4]:
#cal of percentage variation in temperature
#intiation of all variables
# Chapter 2
import math
print"Example 2.4, Page:48  \n \n"
print("Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 4")
t1=0;#temperature at ice point
t2=100;#temperature at steam point
t3=50;#temperature of gas
print("emf equation,e=0.18*t-5.2*10^-4*t^2 in millivolts")
print("as ice point and steam points are two reference points,so")
print("at ice point,emf(e1)in mV")
e1=0.18*t1-5.2*10**-4*t1**2
print("at steam point,emf(e2)in mV")
e2=0.18*t2-5.2*10**-4*t2**2
print("at gas temperature,emf(e3)in mV")
e3=0.18*t3-5.2*10**-4*t3**2
t=((t2-t1)/(e2-e1))*e3
variation=((t-t3)/t3)*100
print("since emf variation is linear so,temperature(t)in degree celcius at emf of 7.7 mV"),round(t,2)
print("temperature of gas using thermocouple=60.16 degree celcius")
print("percentage variation=((t-t3)/t3)*100variation in temperature reading with respect to gas thermometer reading of 50 degree celcius"),round(variation,2)

Example 2.4, Page:48

Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 4
emf equation,e=0.18*t-5.2*10^-4*t^2 in millivolts
as ice point and steam points are two reference points,so
at ice point,emf(e1)in mV
at steam point,emf(e2)in mV
at gas temperature,emf(e3)in mV
since emf variation is linear so,temperature(t)in degree celcius at emf of 7.7 mV 60.16
temperature of gas using thermocouple=60.16 degree celcius
percentage variation=((t-t3)/t3)*100variation in temperature reading with respect to gas thermometer reading of 50 degree celcius 20.31


## example 2.5;pg no:48¶

In [5]:
#cal of absolute zero temperature
#intiation of all variables
# Chapter 2
print"Example 2.5, Page:48  \n \n"
print("Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 5")
print("let the conversion relation be X=aC+b")
print("where C is temperature in degree celcius,a&b are constants and X is temperature in X degree ")
print("at freezing point,temperature=0 degree celcius,0 degree X")
print("so by equation X=aC+b")
X=0;#temperature in degree X
C=0;#temperature in degree celcius
print("we get b=0")
b=0;
print("at boiling point,temperature=100 degree celcius,1000 degree X")
X=1000;#temperature in degree X
C=100;#temperature in degree celcius
a=(X-b)/C
print("conversion relation")
print("X=10*C")
print("absolute zero temperature in degree celcius=-273.15")
X=10*-273.15
print("absolute zero temperature in degree X="),round(X,2)

Example 2.5, Page:48

Engineering Thermodynamics by Onkar Singh,Chapter 2,Example 5
let the conversion relation be X=aC+b
where C is temperature in degree celcius,a&b are constants and X is temperature in X degree
at freezing point,temperature=0 degree celcius,0 degree X
so by equation X=aC+b
we get b=0
at boiling point,temperature=100 degree celcius,1000 degree X
conversion relation
X=10*C
absolute zero temperature in degree celcius=-273.15
absolute zero temperature in degree X= -2731.5