# Chapter 3 : The mechanical equivalent of heat¶

## Example 3.1 pageno : 44¶

In :
# Variables
m = 20;			#calorimeter of water equivalent in gm
n = 1030;			#weight of water in gm
a = 10;			#weight of each paddle in kg
s = 80;			#dismath.tance between paddles in m
g = 980;			#accelaration due to gravity in cm/sec**2

# Calculations
E = (p*a*1000*g*s*100);			#potential energy in dyne cm
T = (E)/(1050*4.18*10**7);			#rise in temperature in deg.C

# Result
print 'the rise in temperature of water is %3.2f deg.C'%(T)

the rise in temperature of water is 3.57 deg.C


## Example 3.2 pageno : 45¶

In :
# Variables
cp = 0.1;			#specific heat of copper in kj/kg-K
w = 120;			#weight of copper calorimeter in gm
a = 1400;			#weight of paraffin oil in gm
cp1 = 0.6;			#specific of parafin oil in kj/kg-K
b = 10**8;			#force to rotate the paddle in dynes
T = 16;			#rise in temperature in deg.C
n = 900;			#no.of revolutions stirred
pi = 3.14;			#value of pi

# Calculations
c = 2*pi*b;			#work done by a rotating paddle per rotation in dyne cm per rotation
d = c*n;			#total work done in dyne cm
hc = w*cp*16;			#heat gained by calorimeter in calories
hp = a*cp1*16;			#heat gaained by paraffin oil in calories
J = d/(hc+hp);			#mecanical equivalent of heat in erg/cal

# Result
print 'mecanical equivalent of heat is %.2e erg/cal'%(J)

mecanical equivalent of heat is 4.15e+07 erg/cal


## Example 3.3 pageno : 45¶

In :
# Variables
cp = 0.12;			#specific heat of iron in kj/kg-K
m = 25;			#mass of iron in lb
h = 0.4;			#horse power developed in 3 min
t = 3;			#time taken to develop the horse power in min
T = 17;			#raise in temp in deg.C

# Calculations
w = h*33000*t;			#total work done in ft-lb
H = m*cp*T;			#aount of heat developed in B.Th.U
J = (w)/H;			#the value of mechanical equivalent of heat

# Result
print 'the mechanical equivalent of water is %3.1f ft-lb/B.Th.U'%(J)

the mechanical equivalent of water is 776.5 ft-lb/B.Th.U


## Example 3.4 pageno : 45¶

In :
# Variables
n = 2.;			#no.of lead blocks
m = 210.;			#mass of each lead block in gm
v = 20000.;			#velocity of block relative to earth in cm/sec
J = 4.2*10**7;			#mechanical equivalent of heat in ergs/calorie
cp = 0.03;			#specific heat of lead in kj/kg-K

# Calculations
E = (m*v**2)/2;			#kinetic energy of each block in ergs
E2 = n*E;			#total kinetic energy in ergs
T = E2/(J*m*n*cp);			#mean rise in temperature in T

# Result
print 'the mean rise in temperature is %3.1f deg.C'%(T)

the mean rise in temperature is 158.7 deg.C


## Example 3.5 pageno : 45¶

In :
# Variables
h = 150;			#height froom which ball fallen in ft
cp = 0.03;			#specific heat of lead in kj/kg-K
J = 778;			#mechanical equivalent of heat in ft lb/B.Th.U

# Calculations
#work done in falling is equal to heat absorbed by the ball
T = 160./(J*cp)*(5./9);			#the raise in temperature in T

# Result
print 'the raise in temperature is %3.1f deg.C'%(T)

the raise in temperature is 3.8 deg.C


## Example 3.6 pageno : 46¶

In :
import math
# Variables
w = 26.6;			#work done one horse in to raise the temperature in lb
T1 = 32.;			#temperature at initial in deg.F
T2 = 212.;			#temperature at final in deg.F
t = 2.5;			#time to raise the tmperature in hrs
p = 25.;			#percentage of heat lossed

# Calculations
#only 75% of heat is utillised
x = w*180.*100.*778./((100-p)*150);			#the rate at which horse worked

# Result
print 'the rate at which horse worked is %3.0f ft-lb wt/min'%(x)

the rate at which horse worked is 33112 ft-lb wt/min


## Example 3.7 pageno : 46¶

In :
# Variables
l = 100.;			#length of glass tube in cm
m = 500.;			#mass of mercury in glass tube in gm
n = 20.;			#number of times inverted i succession
cp = 0.03;			#specific heat of mercury in cal/gm/deg.C
J = 4.2;			#joule's equivalent in j/cal
g = 981.;			#accelaration due to gravity in cm/s**2

# Calculations
PE = m*g*l;			#potential energy for each time in ergs
TE = PE*n;			#total loss in ergs
T = TE/(m*cp*J*10**7);			#rise in temperature in deg.C
#if T is the rise in temperature,then heat devoloped is m*cp*T

# Result
print 'the rise in temperature is %3.2f deg.C'%(T)

the rise in temperature is 1.56 deg.C


## Example 3.8 page no : 46¶

In :
# Variables
d = 0.02;			#diameter of the copper wire in cm
i = 1;			#current in amp
T = 100;			#maximum steady temperature in deg.C
r = 2.1;			#resistance of the wire in ohm cm
J = 4.2;			#mechanical equivalent of heat in j/cal
a = 3.14*d**2/4;			#area of the copper wire in sq.cm
a2 = 1;			#area of the copper surface in sq.cm

# Calculations
l = 1/(2*3.14*d/2);			#length corresponding to the area in cm
R = r*l/a;			#resistance of the copper wire in ohm
w = R*a2**2;			#work done in joule
h = w/J;			#heat devoleped in cal

# Result
print 'the heat developed is %.f calories'%(round(h,-1))

the heat developed is 25360 calories


## Example 3.9 pageno: 47¶

In :
import math

# Variables
h = 10000;			#vertical height of water fall in cm
v = 5;			    #volume disharged per sec in litres
J = 4.18;			#joule's constant in j/cal
g = 981;			#accelaration due to gravity in cm/sec**2

# Calculations
m = v*1000;			#mass of water disharged per sec in gm
w = m*h*g;			#work done in falling through 100m in erg
H = (v*10**7 *g)/(J*10**7);	#quantity of heat produced in cal
T = H/m;			#rise in temperature in deg.C

# Result
print 'the quantity of heat produced is %3f cal  \
\nthe rise in temperature is %3.2f deg.C'%(H,T)

print "Note : Answer for part A in book is wrong. Please calculate manually."

the quantity of heat produced is 1173.444976 cal
the rise in temperature is 0.23 deg.C
Note : Answer for part A in book is wrong. Please calculate manually.


## Example 3.10 page no : 47¶

In :
# Variables
cp = 0.03;			#specific heat of lead in kj/kg.k
v = 10000;			#initial velocity of bullet in cm/sec
J = 4.2*10**7;			#joules constant in ergs/cal

# Calculations
ke = (v**2)/2;			#kinetic energy of the bullet per unit mass in (cm/sec)**2
T = ke*95/(cp*J*100);			#rise in temperature in deg.C

# Result
print 'the rise in temperature is %3.1f deg.C'%(T)

the rise in temperature is 37.7 deg.C


## Example 3.11 page no : 47¶

In :
# Variables
h = 5000.;			#height of the niagara falls in cm
J = 4.2*10**7;		#joules constant in ergs per cal
g = 981;			#accelaration due to gravity in cm/sec**2

#CALCULATIONS
w = h*g;			#work done per unit mass in ergs/gn
T = w/J;			#rise in temperature in deg.C

# Result
print 'the rise in temperature is %3.2f deg.C'%(T)

the rise in temperature is 0.12 deg.C


## Example 3.12 page no : 48¶

In :
import math

# Variables
E1 = 3.75;			#potential difference in v
E2 = 3.;			#potential differnce in v
i1 = 2.5;			#current in amp
i2 = 2;			    #current in amp
T = 2.7;			#the rise in temperature of the water in deg.C
m1 = 48.;			#water flow rate at 3 volts in gm/min
m2 = 30.;			#water flow rate at 3.75volts in gm/min
s = 1;			    #specific heat of the water kj/kg-K

# Calculations
J = (E1*i1-E2*i2)/(s*T*(m1-m2)/60);			#the mechanical equivalent in j/cal

# Result
print 'the mechanical equivalent is %3.3f j/cal'%(J)

the mechanical equivalent is 4.167 j/cal


## Example 3.13 page no : 48¶

In :
# Variables
R = 64*10**7;			#mean radius of the earth in cm
cp = 0.15;			#specific heat of earth in kj/kg-K
J = 4.2*10**7;			#joules consmath.tant in erg/cal

# Calculations
i = 2./5*R**2;			#moment of inertia of the earth per unit mass in joules
w = (2*3.14)/(24*60*60);			#angular velocity of the earth in rad/sec
T = (i*w**2)/(2*J*cp);			#rise in temperature in deg.C

# Result
print 'the rise in the temperature is %.1f deg C'%(T)

the rise in the temperature is 68.7 deg C


## Example 3.14 page no : 49¶

In :
# Variables
cp = 1.25;			#specific heat of helium inkj/kg-K
v = 1000;			#volume of the gas in ml
w = 0.1785;			#mass of the gas at N.T.P in gm
p = 76*13.6*981;	#pressure of the gas at N.T.P in dynes
T = 273;			#temperature at N.T.P in K

# Calculations
V = 1000/w;			#volume occupied by the 1gm of helium gas in cc
cv = cp/1.66;		#specific heat at constant volume it is monatomuc gas kj/kg-K
r = p*V/T;			#gas constant in cm**3.atm./K.mol
J = r/(cp-cv);		#mechanical equivalent of heat in erg/cal

# Result
print 'the mechanical equivalent of heat is %.2e ergs/calories'%(J)
print "Note: answer slightly different because of rounding error."

the mechanical equivalent of heat is 4.19e+07 ergs/calories
Note: answer slightly different because of rounding error.


## Example 3.15 pageno : 49¶

In :
# Variables
n = 1./273;     			#coefficent of expaaansion of air
a = 0.001293;	    		#density of air in gm/cc
cp = 0.2389;		    	#specific heat at consmath.tant pressure in kj/kg.K
p = 76*13.6*981;			#pressure at 0 deg.C in dynes

# Calculations
J = (p*n)/(a*(cp-(cp/1.405)));			#mechanical equivalent of heat

# Result
print 'mechanical equivalent of heat is %.2e ergs/cal'%(J)
print "Note: answer slightly different because of rounding error."

mechanical equivalent of heat is 4.17e+07 ergs/cal
Note: answer slightly different because of rounding error.


## Example 3.16 pageno : 49¶

In :
import math
# Variables
r = 120./60;			#rate of flow of water in gm/sec
T1 = 27.30;			#temperature at initial in deg.C
T2 = 33.75;			#temperature at final in deg.C
v = 12.64;			#potential drop in volts
s = 1.; 			#specific heat of water in kj/kg-K
i = 4.35;			#current through the heating element in amp

# Calculations
J = (v*i)/(r*s*(T2-T1));			#the mechanical equivalent of heat in joule/calorie

# Result
print 'the mechanical equivalent of heat is %3.2f j/cal'%(J)
print "Note: answer slightly different because of rounding error."

the mechanical equivalent of heat is 4.26 j/cal
Note: answer slightly different because of rounding error.


## Example 3.17 page no : 50¶

In :
# Variables
cp = 6.865;			#molar specific heat of hydrogen at consmath.tant pressure in kj/kg-K
cv = 4.880;			#molar specific heat of hydrogen at consmath.tant volume in kj/kg-K
p = 1.013*10**6;			#atmospheric pressure in dynes/cm**2
v = 22.4*10**3;			#gram molar volume in ml
T = 273;			#temperature at N.T.P in kelvins

# Calculations
J = (p*v)/(T*(cp-cv));			#mechanical equivalent of heat

# Result
print 'the mechanical equivalent of heat is %.2e ergs/cal'%(J)

the mechanical equivalent of heat is 4.19e+07 ergs/cal


## Example 3.18 page no : 50¶

In :
import math
# Variables
v = 1000.;			#volume of hydrogen in ml
t = 273.;			#tempature of hydrogen in kelvin
p = 76.;			#pressure of hydrogen in mm of hg
w = 0.0896;			#weigh of hydrogen in gm
cp = 3.409;			#specific heat of hydogen in kj/kg-K
cv = 2.411;			#specific heat of hydrogen in kj/kg-K
g = 981.;			#accelaration due to gravity in cm/sec**2
a = 13.6;			#density of mercury in gm/cm**2

# Calculations
J = (p*v*g*a)/(w*t*(cp-cv));			#mechanical equivalent of heat in ergs/cals

# Result
print 'mechanical equivalent of heat is %.2e ergs/calorie'%(J)

mechanical equivalent of heat is 4.15e+07 ergs/calorie


## Example 3.19 page no : 50¶

In :
# Variables
cp = 0.23;			#specific heat at constant pressure in kj/kg-K
a = 1.18;			#density of air in gm/lit
J = 4.2*10**7;			#mechanical equivalent of heat in ergs/cal
t = 300;			#temperature of air in kelvin
p = 73*13.6*981;			#pressure of air in dynes
#cp-cv = (r/J) = pv/(tj)

#CALCULATON
cv = cp-(p*1000/(a*t*J));			#specific heat at constant volume in calories

# Result
print 'the specific heat at constant volume is %.4f calories'%(cv)
print "Note: answer slightly different because of rounding error."

the specific heat at constant volume is 0.1645 calories
Note: answer slightly different because of rounding error.


## Example 3.20 pageno : 51¶

In :
# Variables
t1 = 0;			#temperature of water in deg.C
t2 = 0;			#temperature of ice in deg.C
J = 4.18*10**7;			#the joules thomson coefficent in erg/cal
l = 80;			#latent heat og fusion kj/kg
g = 981;			#accelaration due to gravity in cm/sec**2

# Calculations
h = l*J/(15*g);			#height from which ice has fallen

# Result
print 'the height from which ice has fallen is %.2e cm'%(h)

the height from which ice has fallen is 2.27e+05 cm


## Example 3.21 page no : 51¶

In :
# Variables
T = 80;			#temperature of bullet in deg.C
cp = 0.03;			#specific heat of lead in kj/kg-K
J = 4.2;			#mechanical equivalent of heat in j/cal

# Calculations
h = T*cp;			#heat developed per unit mass in calorie
v = (J*10**7*h*2/0.9)**0.5;			#velocity of bullet in cm/sec

# Result
print 'the velocity of bullet is %.1e cm/sec'%(v)

the velocity of bullet is 1.5e+04 cm/sec


## Example 3.22 pageno : 51¶

In :
# Variables
w = 5.0;			#weight of lead ball in lb
cp = 0.032;			#specific heat of lead in Btu/lbdeg.F
h = 50;			#height at which ball thrown in feets
v = 20;			#vertical speed in ft/sec
g = 32;			#accelararion due to gravity in ft/sec**2

# Calculations
u = (v**2)+2*g*h
ke = (w/2*(u));			#kinetic energy of the ball at ground
T = ke/(2*32*778*w*cp);			#rise of temperature in deg.F

# Result
print 'the rise in temperature is %.1f deg.F'%(T)

the rise in temperature is 1.1 deg.F