Welcome to Chem Zipper.com......: HEAT, HEAT EXCHANGE AND HEAT CAPACITY(Q)

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Tuesday, September 25, 2018


Heat is defined as the energy that flow into or out of the system because of a difference in temperature between system and surrounding.

Note; Work is more organised way of energy transfer as compared to hear exchange.

IUPAC  Sign convention of Heat: sign of heat will negative (-Ve) if heat is released by the system given by system while sign of heat will be positive (+Ve) if heat is given to the system.
(i) qV = nCvdT (for constant volume process)
(ii) qp = nCpdT (for constant pressure process)
(iii) Cp,m – Cv,m = R
(iv) Cv & Cp depends on temperature even for an ideal gas.( C = a + bT + cT2 .....)
(v) It is a path function
Cv, Cp are heat capacity of system and Cv,m, Cp,m are heat capacity of one mole system at constant  volume and pressure respectively.

“Exchange of heat and work(P–V) in between system and surrounding always
   occur through boundry of system”

   Note: heat exchange can be measured with the help of Heat Capacity.

 We know that usually on increasing in temperature is proportional to the heat transfer
                               q=coefficient x ∆T
The coefficient depend on the size, composition and nature of the system so we can also write it as   
 where C is called Heat capacity
The heat capacity (C):
                                     q=C∆T  or
                                   C= q/∆T      unit- J/K
                            If ΔT=10=1K
                                Then C=q
It is equal to amount of heat needed to raise the temperature of the sample of any substance by one degree Celsius (or Kelvin).

heat Capacity depend  on quantity, nature as well as physical state of the system. And the heat capacity is extensive. it may be made intensive as specific heat capacity
Specific Heat Capacity(Cs ): 
                                     q=Csm∆T  or
                                   Cs= q/m∆T      
                            If ΔT=10=1K and m = 1g
                                Then Cs=q
It is equal to amount of heat required to raise the temperature of 1gm substance by one degree centigrade .it is intensive properties.

Molar Heat Capacity(Cm ):
                                     q=Cmn∆T  or
                                   Cm= q/n∆T      
                            If ΔT=10=1K and n=1mole

                                Then Cm=q
It is equal to amount of heat required to raise the temperature of 1 Mole substance by one degree centigrade

EXAMPLE(1): The latent heat of fusion of ice at 0ºCis 80 cal/gm the amount of heat needed to convert 200 gm ice into water at 0ºC is ?
 (A) 80 cal (B) 16000 J (C) 16000 cal (D) 1600 cal

SOLUTION: Ans (C)  q = m.L = 200 × 80 = 16000 cal

EXAMPLE(2): Calculate the amount of heat required to raise the temperature of 50 gm water from 25ºC to 55ºC.Specific heat capacity of water = 4.2 J/ºC-gm.
(A) 126 J    (B) 210 J    (C) 6300 J   (D) 1500 J

  q = m.s.ΔT = 50 × 4.2 × (55 – 25) = 6300 J

EXAMPLE (3): Five moles of a monatomic ideal gas is heated from 300K to 400K at constant pressure. the amount of heat absorbed is :
(A) 500 cal   (B) 1500 cal   (C) 2500 cal   (D) 2500 J

SOLUTION: Ans. (C)     
qp = CpΔT = n.Cp,m ΔT
    = 5 × 5/2× (400 – 300) = 2500 cal

EXAMPLE (4):  2 moles of an ideal gas absorbs 720 cal heat when heated from 27ºC to 87ºC, at constant volume. 'ɤ' for the gas is :
(A) 1.5 (B) 1.4 (C) 1.6 (D) 1.33

 qv = n.Cv,m. ΔT
Cv,m =qv/nΔT
       = 720/2x(87 – 27)
       = 6 cal/K-mol
Now, r = 1 +R/Cv,m
           = 1 + 2/6
           = 1.33

EXAMPLE (5): 500 gm ice at 0ºC is added in 2000 gm water at tºC. If the final temperature of system is 0ºC, then the value of 't' is (latent heat of fusion of ice = 80 cal/gm and specific heat capacity of water=cal/gm-ºC)
(A)   20    (B)  40      (C) 10     (D)     2

Heat lost by water = heat gained by ice
or, (m.s. ΔT)water = (m.L)ice
or, 2000 × 1 × (t – 0) = 500 × 800
Δ t = 20ºC

EXAMPLE(6): What is the heat in Joules required to raise the temperature of 25 grams of water from 0 °C to 100 °C? What is the heat in calories?
(Given: specific heat of water = 4.18 J/g·°C)

SOLUTION: Use the formula   q = mcΔT
= heat energy
m = mass
c = 
specific heat
ΔT = change in temperature

q = 25gx4.18 J/g·°x(100 °C - 0 °C)
q = 25gx4.18 J/g·°Cx(100 °C)
q = 10450 J
 We know 1 Calorie=4.18 J
So  10450 J in Calorie   = 10450/4.8=2500 calorie 



Case –(1)  when Cm is constant
              Q= nCm (T2-T1)
Case- (2) Cm = f(T)
                 Cm = a + bT+ cT2 +……..

Case- (3) The theoretical value of Cvm and Cpm  for Ideal gas can determined by using degree of freedom.


(1): The heat capacity Of any system should depend upon temperature because by increasing    temperature  of system different degree of freedom get excited.
(2): When temperature approaches zero then heat absorbed by the solid mainly converted into vibration potential energy of molecule resulting in very small increase in temperature, hence ‘C’ increases sharply with increase in temperature.
                   Normally C Directly proportional T3
 (3): When the temperature at Melting point of solid ,then heat capacity becomes nearly  constant  for solid elements.
                    Molar heat capacity= 6.4 Cal/K mole
              Or specific heat capacity x atomic weight=6.4 (Dulong and petite’s law)
(4): Exactly at melting point, the heat capacity become infinite as  ΔT=0
(5): the heat capacity of liquid is greater than that of solid because of rotational degree of freedom also excited.
(6): In liquid heat capacity also depend upon temperature and also infinite at boiling point.
(7): the heat capacity of gas become less than liquid because all vibrational and rotational degree of freedom converted into translational degree of freedom.
(8): the  heat capacity of gases depend upon their atomicity.
(9): If the heat capacity depends upon temperature.
(10); As heat (q) is path function, any substance may have infinite heat capacity.
         Example for any substance.
         Isothermal process = infinite
         Adiabatic process   = 0
         Isobaric process     = Cp
         Isochoric process   = Cv

      Normally ,we use Cp and Cv  value  as characteristic of substance.


It is equal to number of modes of energy transfer when a gaseous molecule undergoes collision.                      OR

        It represent the number of independent modes to describe the molecular motion.  

Total degree of freedom = 3N (Where N  is Number of atom in molecule)

1- Translational degree of freedom is 3 (three) always for mono,di and tri atomic molecule.
2- Rotational degree of freedom is zero for mono atomic gas,2 (two) for diatomic molecules and    3 (three) for triatomic molecule

3-Vibrational degree of freedom is also zero for mono atomic gas and 1(one) diatomic gas molecule  and for polyatomic gases VDOF is calculated individually.( fvib= 3N- ftrans+ frot)

Total degree of freedom:=  ftrans + f rot + f vib     and     fvib= 3N- ftrans + f rot    

Molecules                N                      TDF
He                             1                         3
O2                                    2                         6
CO2                           3                         9
NH3                            4                        12
PCl5                           6                        18


Monoatomic       Diatomic       Triatomic (linear)       Triatomic (Non linear)
f total =3                ftotal  =6          ftotal =9                      f total=9
f trans=3                  ftrans =3            ftrans=3                     ftrans=3
f rot  =0                      frot     =2           frot    = 2                    frot    =3
f vib  =0                     fvib     =1           fvib    = 4                        fvib   =3

                                  Q =n CmdT

                                QV=n CvmdT

 By FLOT dq=dU+dW     and  at constant volume  dW=0   so   dQv=dU

                         Hence    Cvm= (dU/dT)v

Average energy associate with each molecule per degree of freedom is U= 1/2KT  (where K is Boltz’s man constant.

Let degree of freedom is = f   then U is U=1/2fkT

             And U=1/2fkTNper molecule  we know  kNA=R

                     U=1/2fRT  and   dU/dT=1/2fR

             And  dU/dT=Cv      hence  Cv=1/2fR

Cv=1/2ftransR +1/2frotR  (Where Vib degree inactive in chemistry)

       For ideal gas Cpm-Cvm=R  and  Gama= Cpm/Cvm

Adiabatic exponent :Adiabatic exponent (Gama) for a mixture of gas with different heat capacity is defined as :
where n1, n2 ........................ are moles of different gases

Example:Calculate change in internal energy of 10 gm of H2 ,when it's state is changed from(300K, 1Atm) to (500 K, 2Atm) ?
Solution: For ideal gas
Cv for H2 (diatomic) in low temperature range will be 5R as vibrational part is not included.

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