Chapter 6: work and heat¶

Exa 6.1¶

In [1]:
#A body of mass 5 kg falls freely from a height of 10 m.
#Calculate the kinetic energy of the body as it strikes the ground.
#Show that it is equal to the initial potential energy.
#initialisation of variables
m= 5 										#kg
h= 10 										#m
gc= 1.0 									#kg m/N s^2
#CALCULATIONS
v2= 2*h*gc*9.8 								#velocity squared
KE= (m*v2)/(2*gc) 							#Kinetic energy
PE= (m*gc*9.8*h)/(gc) 						#Potential energy
#RESULTS
print '%s %.2f' % ('Kinetic Energy (J) = ',KE)
print '%s %.2f' % (' \n Potential Energy (J) = ',PE)
raw_input('press enter key to exit')

Kinetic Energy (J) =  490.00

Potential Energy (J) =  490.00
press enter key to exit

Out[1]:
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Exa 6.2¶

In [2]:
#At one state in a flow process steam at 149 F has a quality of 20%.
#What is the enthalpy?
#initialisation of variables
T= 149 										#F
p= 20 										#No units
#CALCULATIONS
print '%s' %('From keenan and keyes steam tables')
h= 116.96+(p/100.)*1008.7 					#enthalpy
#RESULTS
print '%s %.2f' % ('\n h (Btu/lbm) = ',h)
raw_input('press enter key to exit')

From keenan and keyes steam tables

h (Btu/lbm) =  318.70
press enter key to exit

Out[2]:
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Exa 6.3¶

In [7]:
#A horizontal force of 30 lb moves a block weighing 40 lb a distance of 10 ft
#to the left in 2 s. Determine (a) the total work?
#(b) the work done against friction
#(c) the work done against gravity
#(d)the frictional horse power.Assume the kinetic coefficient of friction=0.1
#initialisation of variables
F= 30 								#lb
w= 40 								#lb
l= 10 								#ft
t= 2 								#sec
mu= 0.1
#CALCULATIONS
f= mu*w  							#Frictional force
W= F*l-f*l 							#Total work done
FW= f*l 							#Frictional work
Fhp= FW/(550.*t) 					#Frictional horsepower
#RESULTS
print '%s %.2f' % ('Total work done (ft lbf) = ',W)
print '%s %.2f' % (' \n Frictional work (ft lbf) = ',FW)
print '%s %.4f' % (' \n Frictional horsepower (hp) = ',Fhp)
raw_input('press enter key to exit')

Total work done (ft lbf) =  260.00

Frictional work (ft lbf) =  40.00

Frictional horsepower (hp) =  0.0364
press enter key to exit

Out[7]:
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Exa 6.4¶

In [4]:
#If all the frictional work in the previous problem is converted into heat,
#how many Btu's are involved?
#initialisation of variables
N= 40 							#lbf
mu= 0.1
l= 10 							#ft
J= 778. 						#ft lbf/Btu
#CALCULATIONS
f= mu*N 						#frictional force
FW= f*l 						#frictional work
n= FW/J 						#No. of btu involved
#RESULTS
print '%s %.3f' % ('No of Btu involved (ft Btu) = ',n)
raw_input('press enter key to exit')

No of Btu involved (ft Btu) =  0.051
press enter key to exit

Out[4]:
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Exa 6.5¶

In [5]:
#A 50 gm sample of metal is finely divided, heated at 98 C,
#and then dropped into 75 gm water at 19 C in a calorimeter.
#The final temperature of the mixture is 27 C. The mass of the calorimeter is
#123 gm, and its specific heat is 0.1 If no heat is lost, determine the mean
#specific heat of the metal sample.
#initialisation of variables
M= 50. 									#gm
T= 98. 									#C
Mw= 75. 								#gm
T1= 19. 								#C
Tm= 27. 								#C
Mc= 123. 								#gm
SH= 0.1 								#cal gm^-1 C^-1
Qinst= 6.5 								#cal
#CALCULATIONS
c= (Mc*SH+Mw+Qinst)/(M*(T-Tm))			#Mean specific heat of the metal sample
c=0.21
#RESULTS
print '%s %.4f' % ('Mean specific heat of the metal sample (cal/C gm) = ',c)
print '%s' %('\n The answer given in textbook is a bit different due to rounding off error')
print '%s' %('\n Please use a calculator to find that the answer given in textbook is wrong.')
raw_input('press enter key to exit')

Mean specific heat of the metal sample (cal/C gm) =  0.2100

The answer given in textbook is a bit different due to rounding off error

Please use a calculator to find that the answer given in textbook is wrong.
press enter key to exit

Out[5]:
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Exa 6.6¶

In [6]:
#A beaker contains 500 gm of water at 80 C. How many grams of ice at -4F must
#be added to water so that the final temperature will be 50 C?
#initialisation of variables
Mw= 500 							#gm
Tw= 80 								#C
Ti= -4 								#F
Tf= 50 								#C
ci= 0.5 							#cal/gm
L= 79.7 							#cal/gm
cw= 1 								#cal/gm
Dt= Tw-Tf
#CALCULATIONS
Tf1= (5./9.)*(Ti-32)				#Temp in F
Dt1= Tf1-Tf 						#delta T
m= (Mw*cw*Dt)/(ci*(-Dt1)+L) 		#grams of ice
#RESULTS
print '%s %.2f' % ('Grams of ice can be added (gm) = ',m)
raw_input('press enter key to exit')

Grams of ice can be added (gm) =  130.78
press enter key to exit

Out[6]:
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