# Chapter 3 : Calorimetry¶

## Example 3.1 Page No : 49¶

In [1]:
import math
# Input data
T = 5.  # Time taken for a liquid to cool from 80 to 50 degree centigrade in minutes
t11 = 80.  # The initial temperature of the liquid in degree centigrade
t12 = 50.  # The final temperature of the liquid in degree centigrade
t21 = 60.  # If the initial temperature of the liquid in degree centigrade
t22 = 30.  # If the final temperature of the liquid in degree centigrade
ts = 20.  # The temperature of the surrounding in degree centigrade

# Calculations
# The time taken for the liquid to cool from 60 to 30 degree centigrade in
# minutes
T1 = ((math.log((t22 - ts) / (t21 - ts))) /
(math.log((t12 - ts) / (t11 - ts)))) * T

# Output
print 'The time taken for a liquid to cool from 60 to 30 degree centigrade is t = %3.0f minutes ' % (T1)

The time taken for a liquid to cool from 60 to 30 degree centigrade is t =  10 minutes


## Example 3.2 Page No : 55¶

In [2]:
# Input data
dw = 1.  # The density of water in g/cm**3
da = 0.8  # The density of alcohol in g/cm**3
t1 = 100.  # The time taken for the water to cool from 50 to 40 degree centigrade in seconds
t2 = 74.  # The time taken for the alcohol to cool from 50 to 40 degree centigrade in seconds
V = 1.  # Let the volume of either liquid be in cm**3

# Calculations
m = V * dw  # The mass of water in g
M = V * da  # The mass of alcohol in g
w = V  # Water equivalent of each calorimeter in cm**3
# The specific heat of alcohol in calorie/g-K
C = ((((m + w) * t2) / (M * t1)) - (w / M))

# Output
print 'The specific heat of alcohol is C = %3.1f calorie/g-K' % (C)

The specific heat of alcohol is C = 0.6 calorie/g-K


## Example 3.3 Page No : 61¶

In [3]:
import math
# Input data
t = 5.  # Time taken for a body to cool from 60 to 40 degree centigrade in minutes
t11 = 60.  # The initial temperature of the body in degree centigrade
t12 = 40.  # The final temperature of the body in degree centigrade
ts = 10.  # The temperature of the surrounding in degree centigrade

# Calculations
# The constant value for the first case at ts
K = math.log((t12 - ts) / (t11 - ts))
# The temperature after the next 5 minutes in degree centigrade
x = ((math.exp(K)) * (t12 - ts)) + ts

# Output
print 'The temperature after the next 5 minutes is x = %3.0f degree centigrade ' % (x)

The temperature after the next 5 minutes is x =  28 degree centigrade


## Example 3.4 Page No : 63¶

In [4]:
import math
# Input data
T = 4.  # Time taken for a liquid to cool from 70 to 50 degree centigrade in minutes
t11 = 70.  # The initial temperature of the liquid in degree centigrade
t12 = 50.  # The final temperature of the liquid in degree centigrade
t21 = 50.  # If the initial temperature of the liquid in degree centigrade
t22 = 40.  # If the final temperature of the liquid in degree centigrade
ts = 25.  # The temperature of the surrounding in degree centigrade

# Calculations
# The time taken for the liquid to cool from 50 to 40 degree centigrade in
# minutes
T1 = ((math.log((t22 - ts) / (t21 - ts))) /
(math.log((t12 - ts) / (t11 - ts)))) * T

# Output
print 'The time taken for a liquid to cool from 50 to 40 degree centigrade is t = %3.3f minutes ' % (T1)

The time taken for a liquid to cool from 50 to 40 degree centigrade is t = 3.476 minutes


## Example 3.5 Page No : 67¶

In [5]:
import math
# Input data
t = 6.  # Time taken for a liquid to cool from 80 to 60 degree centigrade in minutes
T = 10.  # To find the temperature after the time in minutes
t11 = 80.  # The initial temperature of the liquid in degree centigrade
t12 = 60.  # The final temperature of the liquid in degree centigrade
ts = 30.  # The temperature of the surrounding in degree centigrade

# Calculations
# The constant value for the first case at ts
K = (math.log((t12 - ts) / (t11 - ts))) / (-t)
# The temperature after the next 10 minutes in degree centigrade
x = ((math.exp(-T * K)) * (t12 - ts)) + ts

# Output
print 'The temperature after the next 10 minutes is x = %3.2f degree centigrade ' % (x)

The temperature after the next 10 minutes is x = 42.80 degree centigrade


## Example 3.6 Page No : 69¶

In [6]:
import math
# Input data
t = 5.  # The time taken for a body to cool from 80 to 64 degree centigrade in minutes
t11 = 80.  # The initial temperature of the body in degree centigrade
t12 = 64.  # The final temperature of the body in degree centigrade
t21 = 52.  # The temperature of the body after 10 minutes in degree centigrade
T = 10.  # The time taken for a body to cool from 80 to 52 degree centigrade in minutes
T1 = 15.  # To find the temperature after the time in minutes

# Calculations
# The temperature of the surroundings in degree centigrade
ts = ((t21 * t11) - (t12**2)) / (t11 + t21 - (2 * t12))
# The constant value for the first case at ts
K = (math.log((t21 - ts) / (t12 - ts)))
# The temperature after the next 15 minutes in degree centigrade
x = ((math.exp(K)) * (t21 - ts)) + ts

# Output
print '(1)The temperature of the surroundings is %3.0f degree centigrade \n (2)The temperature after the 15 minutes is %3.0f degree centigrade ' % (ts, x)

(1)The temperature of the surroundings is  16 degree centigrade
(2)The temperature after the 15 minutes is  43 degree centigrade


## Example 3.7 Page No : 76¶

In [7]:
# Input data
t2 = 2.  # The time taken for the liquid to cool from 50 to 40 degree centigrade in minutes
t11 = 50.  # The initial temperature of the liquid in degree centigrade
t12 = 40.  # The final temperature of the liquid in degree centigrade
t1 = 5.  # The time taken for the water to cool from 50 to 40 degree centigrade in minutes
m = 100.  # The mass of water in gms
M = 85.  # The mass of liquid in gms
w = 10.  # Water equivalent of the vessel in gms

# Calculations
# The specific heat of a liquid in calories/g-K
C = (((m + w) * (t2 * 60)) / (M * (t1 * 60))) - (w / M)

# Output
print 'The specific heat of a liquid is C = %3.1f calories/g-K' % (C)

The specific heat of a liquid is C = 0.4 calories/g-K


## Example 3.8 Page No : 79¶

In [8]:
# Input data
V = 22400.  # The volume of One gram molecule of a gas at N.T.P in cm**3
p = 76.  # The pressure in cm of Hg
T = 273.  # The temperature in K

# Calculations
P = p * 13.6 * 981  # The pressure in dynes/cm**2
# The universal gas constant for one gram molecule of a gas in ergs/mole-K
R = (P * V) / T

# Output
print 'The universal gas constant for one gram molecule of a gas is R = %3.4g ergs/mole-K' % (R)

The universal gas constant for one gram molecule of a gas is R = 8.32e+07 ergs/mole-K


## Example 3.9 Page No : 83¶

In [9]:
# Input data
Cp = 0.23  # Specific heat of air at constant pressure
J = 4.2 * 10**7  # The amount of energy in ergs/cal
d = 1.293  # The density of air at N.T.P in g/litre
p = 76.  # The pressure in cm of Hg
T = 273.  # The temperature in K

# Calculations
P = p * 13.6 * 980  # The pressure in dynes/cm**2
V = (1000 / d)  # Volume of one gram of air at N.T.P in cm**3
r = (P * V) / T  # The gas constant for one gram of a gas in ergs/g-K
Cv = Cp - (r / J)  # Specific heat of air at constant volume

# Output
print 'The specific heat of air at constant volume is Cv = %3.4f ' % (Cv)

The specific heat of air at constant volume is Cv = 0.1617


## Example 3.10 Page No : 88¶

In [10]:
# Input data
w = 4.  # The Molecular weight of helium
v = 22400.  # The volume of one gram molecule of a gas at N.T.P in cm**3
p = 76.  # The pressure in cm of Hg
T = 273.  # The temperature in K
J = 4.2 * 10**7  # The amount of energy in ergs/cal

# Calculations
V = (v / w)  # The volume of one gram of helium at N.T.P in cm**3
P = p * 13.6 * 980  # The pressure in dynes/cm**2
r = (P * V) / T  # The gas constant for one gram of a gas in ergs/g-K
C = r / J  # The difference in the two specific heats of one gram of helium

# Output
print 'The difference in the two specific heats of one gram of helium is Cp-Cv = %3.4f' % (C)

The difference in the two specific heats of one gram of helium is Cp-Cv = 0.4947


## Example 3.11 Page No : 94¶

In [11]:
# Input data
V = 25.  # Volume of gasoline consumed by an engine in litres/hour
cv = 6. * 10**6  # The calorific value of gasoline in calories/litre
P = 35.  # The output of the engine in kilowatts

# Calculations
h = V * cv  # Total heat produced by gasoline in one hour in calories
H = h / 3600  # Heat produced per second in cal/s
I = H * 4.2  # Heat produced per second in joules/s or watts
E = ((P * 1000) / I) * 100  # The efficiency in percent

# Output
print 'The efficiency of the engine is %3.0f percent ' % (E)

The efficiency of the engine is  20 percent