Chapter 1 : Heat work and energy

Example 1.1 Page No : 7

In [2]:
# Variables
q = 26.45 			#coloumbs
e = 2.432 			#volts

# Calculations
Q1 = q*e
Q2 = Q1*1.0002*10**7

# Results
print  'Energy expenditure in joules  = %.2f int.joules'%(Q1)
print  ' Energy expenditure in ergs  = %.2e ergs'%(Q2)

Energy expenditure in joules  = 64.33 int.joules
 Energy expenditure in ergs  = 6.43e+08 ergs

Example 1.2 Page No : 9

In [2]:
# Variables
I = 0.565 			#amp
R = 15.43 			#ohms
t = 185 			#secs
Tr = 0.544 			#C

# Calculations
Q1 = I**2*R*t
Q2 = I**2*R*t/Tr

# Results
print  'Heat capacity  = %.f int.joules deg**-1'%(Q2)

Heat capacity  = 1675 int.joules deg**-1

Example 1.3 Page No : 10

In [3]:
# Variables
I = 0.565 			#amp
R = 15.43 			#ohms
t = 185   			#secs
Tr = 0.544 			#C

# Calculations
Q1 = I**2*R*t
Q2 = I**2*R*t/(Tr*4.183)

# Results
print  'Heat capacity  = %.1f calories'%(Q2)

Heat capacity  = 400.4 calories

Example 1.4 Page No : 11

In [4]:
# Variables
v = 1.    			#lit
p = 1. 	    		#atm
h = 76.   			#cm
d = 13.595 			#kg/cm**3
g = 980.66 			#dunes cm**-2
j = 4.18 			#joules

# Calculations
W = v*p
W1 = h*d*g
W2 = W1*10**-4
W3 = W2/j

# Results
print  'Work done in lit-atm  = %.f lit-atm'%(W)
print  ' Work done in dynes  = %.2e dynes cm**-2'%(W1)
print  ' Work done in ergs  = %.2e ergs'%(W2)
print  ' Work done in calories  = %.2f calories'%(W3)

Work done in lit-atm  = 1 lit-atm
 Work done in dynes  = 1.01e+06 dynes cm**-2
 Work done in ergs  = 1.01e+02 ergs
 Work done in calories  = 24.24 calories
In [ ]: