Chapter 2 - Thermodynamics : The first law

Example I1 - Pg 46

In [1]:
#Calculate the change in temperature
#Initialization of variables
Cpm=75 #J/k mol
n=5.55 #mol
q=1 #kJ
#Calculations
deltaT=q*1000/(n*Cpm)
#results
print '%s %.1f %s' %('Change in temperature =',deltaT,'K')

Change in temperature = 2.4 K

Example I4 - Pg 52

In [2]:
#Calculate the change in enthalpy of the sample
#Initialization of variables
n=5.55 #mol
T1=20 #C
T2=80 #K
Cpm=75.29 #J/K mol
#Calculations
H=n*Cpm*(T2-T1)/1000.
#results
print '%s %d %s' %('Enthalpy of the sample changes by',H,'kJ')

Enthalpy of the sample changes by 25 kJ

Example E1 - Pg 45

In [3]:
#Calculate the heat supplied, heat capacity and heat output of the calorimeter
#Initialization of variables
A=1.23 #A
V=12 #V
t=123 #s
Temp=4.47 #C
rise=3.22 #C
#Calculations
q=A*V*t
C=q/Temp
Output= C*rise/1000.
#Results
print '%s %.1f %s' %('heat supplied during calibration =',q,'J')
print '%s %.1f %s' %('\n Heat capacity of the calorimeter =',C,'J/C')
print '%s %.2f %s' %('\n Heat output =',Output,' kJ')

heat supplied during calibration = 1815.5 J

 Heat capacity of the calorimeter = 406.1 J/C

 Heat output = 1.31  kJ

Example E2 - Pg 48

In [2]:
#Calculate the amount by which the person's internal energy falls
#Initialization of variables
work=-622 #kJ
heat=-82 #kJ
#Calculations
U=work+heat
#results
print '%s %d %s' %('The persons internal energy falls by',-U,'kJ')

The persons internal energy falls by 704 kJ