# Chapter 4 - Heat effects¶

## Example: 4.1 Page: 118¶

In [2]:
from __future__ import division
print "Example: 4.1 - Page: 118\n\n"

# Solution

#*****Data*****#
Qp = -327## [kcal]
T = 27 + 273## [K]
R = 2*10**(-3)## [kcal/K mol]
#*************#

# The reaction involved is:
# C2H5OH(l) + 3O2(g) = 2CO2(g) + 3H2O(l)
deltan = 2 - 3#
Qv = Qp - deltan*R*T## [kcal]
print "Value of Qv is %.2f kcal\n"%(Qv)#

Example: 4.1 - Page: 118

Value of Qv is -326.40 kcal



## Example: 4.2 Page: 119¶

In [3]:
from __future__ import division
print "Example: 4.2 - Page: 119\n\n"

# Solution

#*****Data*****#
# Mg + (1/2)O2 = MgO ...............(1)
deltaH1 = -610.01## [kcal]
# 2Fe + (3/2)O2 = Fe2O3 ............(2)
deltaH2 = -810.14## [kcal]
#*************#

# 3Mg + Fe2O3 = 3MgO + 2Fe .........(3)
# Multiplying (1) by 3 and substracting from (2), we get (3):
deltaH = 3*deltaH1 - deltaH2## [kcal]
print "Heat produced in the reaction is %.1f kcal\n"%(deltaH)#

Example: 4.2 - Page: 119

Heat produced in the reaction is -1019.9 kcal



## Example: 4.3 Page: 121¶

In [4]:
from __future__ import division
print "Example: 4.3 - Page: 121\n\n"

# Solution

#*****Data*****#
# 2H2(g) + O2(g) ---------------> 2H2O .....................(1)
deltaH1 = -241.8*2## [kJ/gmol H2]
# C(graphite) + O2(g) =---------> CO2(g) ...................(2)
deltaH2 = -393.51## [kJ/gmol C]
# CH4(g) + 2O2(g) ---------------> CO2(g) + 2H2O(l) ........(3)
deltaH3 = -802.36## [kJ/mol CH4]
#*************#

# For standard heat of formation of methane, (a) + (b) - (c)
# C + 2H2 ------------------------> CH4
deltaHf = deltaH1 + deltaH2 - deltaH3## [kJ/gmol]
print "The standard heat of formation of methane is %.2f kJ/gmol\n"%(deltaHf)#

Example: 4.3 - Page: 121

The standard heat of formation of methane is -74.75 kJ/gmol



## Example: 4.4 Page: 122¶

In [5]:
from __future__ import division
print "Example: 4.4 - Page: 122\n\n"

# Solution

#*****Data*****#
deltaH_C6H12O6 = -1273## [kcal]
deltaH_C2H5OH = -277.6## [kcal]
deltaH_CO2 = -393.5## [kcal]
deltaH_H2O = -285.8## [kcal]
#**************#

# C6H12O6(s) = 2C2H5OH(l) + 2CO2(g) ..........................(A)
deltaH_A = 2*deltaH_C2H5OH + 2*deltaH_CO2 - deltaH_C6H12O6## [kJ]
# C6H12O6(s) + 6O2(g) = 6CO2(g) + 6H2O(l) ...................(B)
deltaH_B = 6*deltaH_CO2 + 6*deltaH_H2O - deltaH_C6H12O6## [kJ]
print "Energy supplied by reaction A is %.1f kJ\n"%(deltaH_A)#
print "Energy supplied by reaction B is %.1f kJ\n"%(deltaH_B)#
if deltaH_A < deltaH_B:
print "Reaction A supplies more energy to the organism\n"
else:
print "Reaction B supplies more energy to the organism\n"

Example: 4.4 - Page: 122

Energy supplied by reaction A is -69.2 kJ

Energy supplied by reaction B is -2802.8 kJ

Reaction B supplies more energy to the organism



## Example: 4.5 Page: 122¶

In [6]:
from __future__ import division
print "Example: 4.5 - Page: 122\n\n"

# Solution

#*****Data*****#
# Zn + S = ZnS ....................................................(A)
deltaH_A = -44## [kcal/kmol]
# ZnS + 3O2 = 2ZnO + 2SO2 .........................................(B)
deltaH_B = -221.88## [kcal/kmol]
# 2SO2 + O2 = 2SO3 ................................................(C)
deltaH_C = -46.88## [kcal/kmol]
# ZnO + SO3 = ZnSO4 ...............................................(D)
deltaH_D = -55.10## [kcal/kmol]
#***************#

# Multiplying (A) by 2 & (D) by (2) and adding (A), (B), (C) & (D)
# Zn + S + 2O2 = ZnSO4
deltaH = 2*deltaH_A + deltaH_B + deltaH_C + 2*deltaH_D## [kcal/kmol for 2 kmol of ZnSO4]
print "Heat of formation of ZnSO4 is %.2f kcal/kmol\n"%(deltaH/2)#

Example: 4.5 - Page: 122

Heat of formation of ZnSO4 is -233.48 kcal/kmol



## Example: 4.6 Page: 124¶

In [7]:
from __future__ import division
print "Example: 4.6 - Page: 124\n\n"

# Solution

#*****Data*****#
# HC : Heat of Combustion
HC_NH3 = -90.6## [kcal]
HC_H2 = -68.3## [kcal]
#*************#

# Heat of combustion of NH3:
# 2NH3 + 3O = N2 + 3H2O ............................ (A)
# Heat of combustion of H2:
# H2 + O = H2O ..................................... (B)
# Multiplying (B) by 3 & substracting from (A), we get:
# 2NH3 = N2 + 3H2 .................................. (C)
# Hf : Heat of Formation
Hf_NH3 = -(2*HC_NH3 - 3*HC_H2)/2## [kcal]
print "Standard Heat of formation of NH3 is %.1f kcal"%(Hf_NH3)#

Example: 4.6 - Page: 124

Standard Heat of formation of NH3 is -11.8 kcal


## Example: 4.7 Page: 125¶

In [8]:
from __future__ import division
from scipy.optimize import fsolve
print "Example: 4.7 - Page: 125\n\n"

# Solution

#*****Data*****#
# HC : Heat of Combustion
HC_C2H2 = -310600# # [cal]
#**************#

# C2H2 + (5/2)O2 = 2CO2 + H2O
Q = -HC_C2H2## [cal]
# The gases present in the flame zone after combustion are carbon dioxide, water vapor and the unreacted nitrogen of the air.
# Since (5/2) mole of oxygen were required for combustion, nitrogen required would be 10 mol.
# Hence the composition of the resultant gas would be 2 mol CO2, 1  ol H2 & 10 mol N2.
# Q = integrate('Cp(T)','T',T,298)#
# On integrating we get:
# Q = 84.52*(T - 298) + 18.3*10**(-3)*(T**2 - 298**2)
#deff('[y] = f(T)','y = Q - 84.52*(T - 298) - 18.3*10**(-3)*(T**2 - 298**2)')#
def f(T):
y = Q - 84.52*(T - 298) - 18.3*10**(-3)*(T**2 - 298**2)
return y

T = fsolve(f,7)## [K]
print "The maximum attainable temperature is %.1f K"%(T)#

Example: 4.7 - Page: 125

The maximum attainable temperature is 2566.2 K


## Example: 4.8 Page: 126¶

In [9]:
print "Example: 4.8 - Page: 126\n\n"

# Solution

#*****Data*****#
Cp_CO2 = 54.56## [kJ/mol K]
Cp_O2 = 35.20## [kJ/mol K]
Cp_steam = 43.38## [kJ/mol K]
Cp_N2 = 33.32## [kJ/mol K]
# 2C2H6(g) + 7O2(g) = 4CO2(g) + 6H2O(g)
deltaH_273 = -1560000## [kJ/kmol]
#************#

# Since the air is 25% in excess of the amount required,the combustion may be written as:
# C2H6(g) + (7/2)O2(g) = 2CO2(g) + 3H2O(g)
# 25% excess air is supplied.
# Since the air contains N2 = 79% and O2 = 21%
# C2H6(g) + 3.5O2(g) + 0.25*3.5O2(g) + (4.375*(79/21))N2 = 2CO2 + 3H2O + 0.875O2 + 16.46N2 .................. (A)
# Considering the reaction (A),
# Amount of O2:
O2 = 3.5 + 3.5*0.25## [mol]
# Amount of N2 required:
N2 = 4.375*(79/21)## [mol]
# Let the initial temperature of ethane and air be 0 OC and the temperature of products of combustion be T OC
# Since heat librated by combustion = heat accumulated by combustion products
Q = -deltaH_273## [kJ/mol K]
T = Q/(2*Cp_CO2 + 3*Cp_steam + 0.875*Cp_O2 + N2*Cp_N2)## [OC]
print "The theoretical temperature of combustion is %d degree Celsius"%(T)#

Example: 4.8 - Page: 126

The theoretical temperature of combustion is 1906 degree Celsius


## Example: 4.9 Page: 129¶

In [10]:
from __future__ import division
print "Example: 4.9 - Page: 129\n\n"

# Solution

#*****Data*****#
T1 = 273## [K]
T2 = 253## [K]
deltaH_273 = 1440## [cal/mol]
Cp = 8.7## [cal/mol]
#**************#

deltaH_253 = deltaH_273 + Cp*(T2 - T1)## [cal/mol]
print "Laten heat of ice at -20 OC is %d cal/mol\n"%(deltaH_253)#

Example: 4.9 - Page: 129

Laten heat of ice at -20 OC is 1266 cal/mol



## Example: 4.10 Page: 129¶

In [11]:
from __future__ import division
print "Example: 4.10 - Page: 129\n\n"

# Solution

#*****Data*****#
T2 = 1273## [K]
T1 = 300## [K]
deltaH_300 = -11030## [cal/mol]
#*************#

# The chemical reaction involved is:
# N2 + 3H2 = 2NH3
# (1/2)N2 + (3/2)H2 = NH3
# deltaH_1273 = deltaH_300 + integrate('Cp_NH3(T) - (1/2)*Cp_N2(T) - (1/2)*Cp_H2(T)','T',1273,300)#
deltaH_1273 = deltaH_300 + quad(lambda T:(6.2 + 7.8*10**(-3)*T - 7.2*10**(-6)*T**2) - (1/2)*(6.45 + 1.4*10**(-3)*T) - (1/2)*(6.94 - 0.2*10**(-3)*T),[1273,300])## [cal]
print "Heat of formation at 1273 K is %d cal"%(deltaH_1273)#

Example: 4.10 - Page: 129

Heat of formation at 1273 K is -11172 cal


## Example: 4.11 Page: 130¶

In [12]:
from __future__ import division
print "Example: 4.11 - Page: 130\n\n"

# Solution

#*****Data*****#
CO2 = 13.4## [percent by volume]
N2 = 80.5## [percent by volume]
O2 = 6.1## [percent by volume]
#*************#

# Basis : 100 cubic m of flue gas.
Vol_N2_flue = N2## [Volume of Nitrogen in flue gas, cubic m]
Vol_O2_flue = O2## [Volume of O2 in flue gas, cubic m]
Vol_Air = N2/0.79## [Volume of air supplied, cubic m]
Vol_O2 = Vol_Air*0.21## [Volume of O2 in air supply, cubic m]
Vol_O2_cumbustion = Vol_O2 - Vol_O2_flue## [Volume of O2 used up in cumbustion of the fuel, cubic m]
Excess_Air = Vol_O2_flue/Vol_O2_cumbustion * 100## [percent of excess air supplied]
print "Percent of excess air supplied is %.1f %%"%(Excess_Air)#

Example: 4.11 - Page: 130

Percent of excess air supplied is 39.9 %