# Chapter 3 : Zeroth Law of Thermodynamics and Temperature Scales¶

## Example 3.1 Page No : 72¶

In :
# Variables
t_c = 303-273;			# in °C

# Calculations and Results
t_f = 9./5* t_c+32;			# in °F
print "When the temperature is 303 K then the thermometer reading in °F is : %.0f"%t_f

T_R = 460 + t_f;			# °R
print "The absolute value of the temperature in Rankine scale in °R is : %.0f"%T_R

When the temperature is 303 K then the thermometer reading in °F is : 86
The absolute value of the temperature in Rankine scale in °R is : 546


## Example 3.2 Page No : 73¶

In :
# Variables
# t_C= t_F or T_K-T_R= -186.52     (i)
# T_R/T_K = 1.68                                (ii)
# From eq (i) and (ii)
T_K= -186.52/(1-1.68);			# temp. in kelvin in K

# Calculations
T_R= 1.68*T_K;			    # in temp. in rankine in °R
t_C= T_K-273.15;			# in °C
t_F= T_R-459.67;			# in °F

# Results
print "Temperature in kelvin is : %.2f"%T_K
print "Temperature in °R is : %.2f"%T_R
print "Temperature in °C is : %.2f"%t_C
print "Temperature in °F is : %.2f"%t_F

Temperature in kelvin is : 274.29
Temperature in °R is : 460.81
Temperature in °C is : 1.14
Temperature in °F is : 1.14


## Example 3.9 Page No : 75¶

In :
import math

# Variables
p0 = 1.86;
p100 = 6.81;
T1=32;
T2= 212;

# Calculations
# Relation of T in terms of p for ice point      T1= a*math.log(p0)+b          (i)
# Relation of T in terms of p for steam point  T2= a*math.log(p100)+b     (ii)
# From eq(i) and (ii)
a= (T2-T1)/math.log(p100/p0);
b= T1-a*math.log(p0);
# The temp at
p=2.5;
T= a*math.log(p)+b;			# in °unit

# Results
print "The temperature at p=2.5 in °unit is : %.3f"%T

The temperature at p=2.5 in °unit is : 73.014


## Example 3.10 Page No : 76¶

In :
from numpy import roots

# Variables
Tp0=0.;			#in °C (at ice point)
Tq0=0.;			#in °C (at ice point)
# Putting these values in relation, we get
a=0.;
Tp100=100.;			#in °C ( at steam point)
Tq100=100.;			#in °C ( at steam point)
# Tp100= b*Tq100+lamda*Tq100**2         (i)
Tp=45.;			# in °C (in oil path)
Tq=43.;			# in °C (in oil path)

# Calculations
# Tp= b*Tq+lamda*Tq**2                          (ii)
b= (Tp100-Tp*Tq100**2/Tq**2)/(Tq100-Tq100**2/Tq);			# From eq (i) and (ii)
lamda= (Tp-b*Tq)/Tq**2;
Tp=20;

#lamda*Tq**2+b*Tq-Tp=0
P= [lamda, b, -Tp];
Tq= roots(P);			# in °C

# Results
print "When P reads 20°C, then the readings of  Q in °C are %.2f C "%(Tq)
print "The realistic value of Tq in °C is : %.2f"%Tq

When P reads 20°C, then the readings of  Q in °C are 18.76 C
The realistic value of Tq in °C is : 18.76