In [1]:

```
# 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)
```

In [2]:

```
# 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)
```

In [3]:

```
# 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)
```

In [4]:

```
# 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)
```

In [5]:

```
# 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)
```

In [6]:

```
# 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)
```

In [7]:

```
# 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)
```

In [8]:

```
# 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)
```

In [9]:

```
# 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)
```