Basic Concept of Specific Heats, Joule’s law and Enthalpy21 min read

What is specific Heat?

The specific heat of a solid or liquid is defined as the amount of heat per unit mass required to a one-degree Celsius temperature rise. for small quantities, we have

 \boxed{ dQ = m.c.dT}

where

  • m : mass of the substance
  • c : Specific Heat
  • dT : Change of temperature in degree Celcius

Specific heats for gases

For the gases, There are two kinds of specific heats are defined i.e.

Specific heat at constant pressure :    c_v

Specific Heat at constant volume :      c_p

So, we have

 dQ = m . c_p . dT          for reversible non flow process at constant pressure.

and

 dQ = m . c_v . dT          for reversible non flow process at constant volume.

The value of  c_p and  c_v is constant for a perfect gas at all pressure and temperature.

so, integrating equations  dQ = m . c_p . dT   and  dQ = m . c_v . dT , we have

flow of heat in a reversible constant pressure process:

 \Rightarrow m. c_p . \left (T_2 - T_1 \right)

flow of heat in a reversible constant volume process:

 \Rightarrow m. c_v . \left (T_2 - T_1 \right)

“In case of real gases,  c_p and  c_v vary with temperature.”

The ratio of Specific Heats

The ratio of specific heat at constant pressure to the specific heat at constant volume is represented by symbol “γ”( Gamma).

So         \gamma = \left (\frac{c_p}{c_v} \right )

Since,  c_p = c_v + R , it is clear that c_p must be greater than c_v for any perfect gas. So the ratio  \gamma =  \frac{c_p}{c_v}  is always greater than 1.

The relationship between two specific Heats

Let us consider a perfect gas being heated at constant pressure from T_1 to T_2 .

According to the closed system or non-flow equation,

 Q = \left( U_2 - U_1 \right) + W

Also, for a perfect gas,

 U_2 - U_1 = m . c_v . \left( T_2 - T_1 \right)

 Q = m . c_v . \left( T_2 - T_1 \right) + W       …..(1)

In a constant pressure process, the work-done by the fluid,

 W = p \left( V_2 - V_1 \right)

 \Rightarrow W = m . R \left(T_2 - T_1 \right)

Since, from the perfect gas equation  p. V = m . R . T , so  p_1. V_1 = m . R . T_1 and  p_2. V_2 = m . R . T_2 and  p_1 = p_2 = p in this case.

So, From equation (1), We have

 Q = m . c_v . \left( T_2 - T_1 \right) + m . R . \left( T_2 - T_1 \right)

 \Rightarrow Q = m \left( c_v + R \right) \left( T_2 - T_1 \right)           …..(a)

But for a constant pressure process,

 Q = m . c_p . \left( T_2 - T_1 \right)           ……(b)

By comparing equations (a) and (b), we have

 m \left( c_v + R \right) \left( T_2 - T_1 \right) = m . c_p . \left( T_2 - T_1 \right)

 \therefore c_v + R = c_p

or  \boxed{c_p - c_v = R }                    ….(2)

Dividing by c_v on both sides

 \Rightarrow \frac{c_p}{c_v} - 1 = \frac{R}{c_v}

 \therefore \gamma -1 = \frac{R}{c_v}

 \Rightarrow \boxed{ c_v = \frac{R}{\gamma - 1} }

Similarly, dividing both sides by c_p, we get

 1 - \frac{c_v}{c_p} = \frac{R}{c_p}

 \Rightarrow 1- \frac{1}{\gamma} = \frac{R}{c_p}

 \therefore \frac{\gamma - 1}{\gamma} = \frac{R}{c_p}

 \boxed{ c_p = \frac{\gamma. R}{\gamma - 1 }}

 

What is Joule’s Law?

Joule’s Law: “The internal energy of the perfect gas is a function of the absolute temperature only.”

i.e.  \boxed{u = f \left(T \right)}

Let 1 kg of a perfect gas be heated at constant volume, according to the non-flow energy equation

 dQ = du + dW

since volume is constant i.e. work-done dW = 0

 \therefore dQ = du

At constant volume for 1 kg of  perfect gas, we have

 dQ = m . c_v . dT

 \therefore dQ = du = c_v. dT

After integrating  u = c_v. T + K

where, K = constant

According to joule’s law  u = f \left( T \right ) that mean internal energy varies linearly with absolute temperature. Internal energy can be made zero at any similar reference temperature. It may be assume that u =0 when T = 0 for a perfect gas so constant K is zero.

i.e., Internal Energy,  u = c_v. T for a perfect gas

or Internal Energy  U = m . c_v . T

for a perfect gas, in any process between state 1 to state 2, we have from equation  U = m . c_v . T

Gain in internal energy,

 \boxed{U_2 - U_1 = m . c_v . \left( T_2 - T_1 \right) }

So the gain of internal energy for a perfect gas between two states for any process, Reversible or irreversible.

What is Enthalpy?

“One of the fundamental quantities which occur invariably in thermodynamics is the sum of internal energy (u) and pressure-volume product (p.v). This sum is called Enthalpy.” Enthalpy is denoted by “h”.

i.e.  h = u + p.v

The enthalpy of a fluid is the property of the fluid since it consists of the sum of property and the product of two properties. Since enthalpy is a property like internal energy, pressure, specific volume and temperature, it can be introduced into any problem whether the process is a flow or a non-flow process.

The total enthalpy of mass, m, of a fluid can be

 \boxed{ H = U + p.V}

where

H = m.h

For a perfect gas:

 h = u + p.v

 \Rightarrow h = c_v . T + R.T

Since  p.v =R.T

 \Rightarrow h = \left( c_v + R \right).T

 \Rightarrow h = c_p .T

Since  c_p = c_v + R

i.e.  \boxed{H = m . c_p .T}

it is assumed that u = 0 at T = 0 then h = 0 at T = 0.

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