# Chapter 4 : Change of State¶

## Example 4.1 Page No : 106¶

In :
# Input data
m = 1000.  # Mass of Ice in gms
Sp = 0.5  # Specific heat of Ice in cal/g-K
t1 = -10  # Initial temperature of Ice in degree centigrade
t2 = 0.  # The final temperature of Ice in degree centigrade
Li = 80.  # Latent heat of fusion of ice in cals per gram
Ls = 540.  # Latent heat of fusion of steam in cals per gram

# Calculations
h1 = m * -t1 * Sp  # Heat required to raise the temperature of Ice in cals
h2 = m * Li  # Heat required to melt ice at 0 degree centigrade in cals
h3 = m * 100  # Heat required to raise the temperature of water from 0 to 100 degree centigrade in cals
h4 = m * Ls  # Heat required to convert water into steam at 100 degree centigrade in cals
T = h1 + h2 + h3 + h4  # Total quantity of heat required in cals

# Output
print 'Total quantity of heat required is %3.0f cals ' % (T)

Total quantity of heat required is 725000 cals


## Example 4.2 Page No : 112¶

In :
# Input data
m = 1.  # Mass of steam in gms
Ls = 537.  # Latent heat of fusion of steam in cal per gram
mi = 100.  # mass of ice in gms
Li = 80.  # Latent heat of fusion of ice in cal per gram

# Calculations
h1 = m * Ls  # Heat given out by one gram of steam when converted from steam into water at 100 degree centigrade in cals
h2 = 1. * 100  # Heat given out by one gram of water when cooled from 100 to 0 degree centigrade in cals
h = h1 + h2  # Total quantity of heat given out by one gram of steam in cals
m = h / Li  # The amount of Ice melted in gms

# Output
print 'The amount of Ice melted is m = %3.2f gms ' % (m)

The amount of Ice melted is m = 7.96 gms


## Example 4.3 Page No : 115¶

In :
# Input data
m = 100.  # Mass of water in gms
tw = 40.  # The temperature of water in degree centigrade
mi = 52.  # Mass of Ice in gms
Lw = 100.  # Latent heat of fusion of water in cals per gram
Li = 80.  # Latent heat of fusion of Ice in cals per gram

# Calculations
h = Lw * tw  # Heat lost by water when its temperature falls from 40 to 0 degree centigrade in cals
hi = mi * Li  # Heat gained by Ice in cals
hg = h  # The amount of heat gained by Ice in cals
ml = (hg / Li)  # The amount of Ice melted in gms
M = mi - ml  # The amount of ice remaining in gms
W = m + (mi - M)  # The amount of water in gms

# Output
print 'The remaining Ice is %3.0f g \n Hence the result will be %3.0f g of Ice and %3.0f g of water at 0 degree centigrade ' % (M, M, W)

The remaining Ice is   2 g
Hence the result will be   2 g of Ice and 150 g of water at 0 degree centigrade


## Example 4.4 Page No : 121¶

In :
# Input data
m = 100.  # Let the mass of water in gms
t = 15.  # Time taken for an electric kettle to heat a certain quantity of water from 0 to 100 degree centigrade in minutes
T = 80.  # Time taken to turn all the water at 100 degree centigrade into steam in minutes
Lw = 100.  # Latent heat of fusion of water in cals per gram

# Calculations
h1 = m * Lw  # Heat required to raise its temperature from 0 to 100 degree centigrade in cals
h2 = h1  # Heat produced by electric kettle in 15 minutes in cals
h3 = h2 / 15  # Heat produced by electric kettle in 1 minute in cals
h4 = h3 * 80  # Heat produced by electric kettle in 80 minutes in cals
L = h4 / m  # Latent heat of steam in cal/g

# Output
print 'The latent heat of steam is L = %3.2f cal/g ' % (L)

The latent heat of steam is L = 533.33 cal/g


## Example 4.5 Page No : 123¶

In :
# Input data
m = 50.  # Mass of water in gms
t1 = 15.  # Initial temperature in degree centigrade
t2 = -20  # Final temperature in degree centigrade
Sp = 0.5  # Specific heat of Ice in cal/g-K
Li = 80.  # Latent heat of fusion of Ice in cals per gram

# Calculations
h1 = m * 1 * t1  # Heat removed in cooling water from 15 to 0 degree centigrade in cal
h2 = m * Li  # Heat removed in converting water into Ice at 0 degree centigrade in cal
h3 = m * Sp * -t2  # Heat removed in cooling ice from 0 to -20 degree centigrade in cal
H = h1 + h2 + h3  # Total heat removed in one hour in cal
H1 = H / 60  # Heat removed per minute in cal/minute

# Output
print 'The Quantity of heat removed per minute is %3.1f cal/minute ' % (H1)

The Quantity of heat removed per minute is 87.5 cal/minute


## Example 4.6 Page No : 128¶

In :
# Input data
M = 20.  # Mass of the substance in g
t = 100.  # The temperature of the substance in degree centigrade
a = 1. / 100  # Area of cross section in cm**2
l = 5.  # The length of the coloumn through which liquid moves in cm
V1 = 1000.  # The volume of water in cm**3
V2 = 1090.  # The volume of Ice from the volume of water on freezing in cm**3
Li = 80.  # Latent heat of Ice in cals per gram

# Calculations
V = V2 - V1  # The decrease in volume of Ice in cm**3
Vi = V / 1000  # The decrease in volume when one gram of ice melts in cm**3
v = l * a  # Decrease in volume in cm**3
# Specific heat of the substance incal/g degree centigrade
S = (Li * v) / (Vi * M * t)

# Output
print 'The specific heat of the substance is %3.3f cal/g.degree centigrade ' % (S)

The specific heat of the substance is 0.022 cal/g.degree centigrade


## Example 4.7 Page No : 132¶

In :
# Input data
M = 27.  # The mass of the substance in g
t = 100.  # The temperature of the substance in degree centigrade
a = 3. / 100  # Area of cross section in cm**2
l = 10.  # The length of the coloumn through which liquid moves in cm
Li = 80.  # Latent heat of Ice in cals per gram
V1 = 1000.  # The volume of water in cm**3
V2 = 1090.  # The volume of Ice from the volume of water on freezing in cm**3

# Calculations
v = l * a  # Decrease in volume in cm**3
V = V2 - V1  # The decrease in volume of Ice in cm**3
Vi = V / 1000  # The decrease in volume when one gram of ice melts in cm**3
# Specific heat of the substance incal/g degree centigrade
S = (Li * v) / (Vi * M * t)

# Output
print 'The specific heat of the substance is %3.3f cal/g.degree centigrade ' % (S)

The specific heat of the substance is 0.099 cal/g.degree centigrade


## Example 4.8 Page No : 134¶

In :
# Input data
t = 16.5  # The temperature of air in degree centigrade
d = 6.5  # The dew point in degree centigrade
s1 = 7.05  # S.V.P at 6 degree centigrade in mm
s2 = 7.51  # S.V.P at 7 degree centigrade in mm
s3 = 13.62  # S.V.P at 16 degree centigrade in mm
s4 = 14.42  # S.V.P at 17 degree centigrade in mm

# Calculations
s5 = (s1 + s2) / 2  # S.V.P at 6.5 degree centigrade in mm
s6 = (s3 + s4) / 2  # S.V.P at 16.5 degree centigrade in mm
R = (s5 / s6) * 100  # Relative humidity of air in percent

# Output
print 'The percentage relative humidity of air is R.H = %3.1f percent ' % (R)

The percentage relative humidity of air is R.H = 51.9 percent


## Example 4.9 Page No : 140¶

In :
# Input data
R = 52.  # The relative humidity of air in percent
t = 20.  # The temperature of air in degree centigrade
s1 = 17.5  # S.V.P of water at 20 degree centigrade in mm
s2 = 9.2  # S.V.P of water at 10 degree centigrade in mm
s3 = 8.6  # S.V.P of water at 9 degree centigrade in mm

# Calculations
s4 = (R / 100) * s1  # S.V.P at dew point in mm
s5 = s2 - s3  # S.V.P for 1 degree centigrade difference in mm
# The dew point temperature in degree centigrade
d = 9. + ((s4 - s3) / (s2 - s3))

# Output
print 'The dew point temperature is %3.2f degree centigrade ' % (d)

The dew point temperature is 9.83 degree centigrade