# Analysis of Thermodynamic Equations

Equations                          Analysis

(i) dQ = dE + dW                  Applicable for any process.

(ii) dQ = dU +dW -7          Applicable for any process undergone by a closed stationarsystem.

(iii) dQ = dU + PdV           Applicable for closed system when PdV is the only work present.

True for reversible process.

(iv) dQrev = TdS               True for reversible process

(v) TdS = dU + PdV          Applicable for any process in closed system.

(vi) TdS = dH – VdP        Applicable for any closed system.

SOLVED EXAMPLES

Example 8.1. In a process friction causes the temperature rise of water from 40°C to 43°C. The process is adiabatic what is the change of entropy per kg of water flow at constant volume? [C (water) = 4.187 KJ/kg k]

Solution:               dQrev = TdS

or                                       dS= dQ / T = mc d T/ T

or

= 1 x 4187 x in ( 43 + 273 / 40 + 273)

S2 – S1 = 0.04 kJ / k.

Example 8.2. Two kgs of water at 300 K is brought into contact with heat reservoir at 400 K.

When the water has reached 400 K, find the entropy change for (a) water (b) heat reservoir (c)

universe.

Solution.

= mc u In T2 / T1 = 2 x 4.187 In 400 / 300

= 2.41 KJ/k

(b) Heat                                   Q = mcv (T2 – T1)

= 2 x 4.187 (400 – 300)

= 837.4 KJ

For reservoir

( ΔS)res= – Q / T = – 837.4 / 400 = – 2.09 KJ / K

(c) Change of entropy of the universe

= 2.41 – 2.09

= 0.32 KJ /K.

Example 8.3. Two kgs of ice at -10°C is exposed to the atmosphere which is at 25°C. The ice

melts and comes into thermal equilibrium with the atmosphere.

Determine the entropy increase of the universe.

[ice= 2.093, Lice= 333.3, C water = 4.187]

Solution.

Heat absorbed of ice at – 10°C to become 0°C ice = Q1 = mice (0 + 10)

= 2 x 2.093 x 10 = 41.86 KJ (at solid phase)

Heat absorbed of 0°C ice to become 0°C water Q2 = 2 x 333.3

= 666.6 KJ (transition phase)

Heat absorbed of 0°C water to become 25°C water

Q3 = 2 x4.187 x (25- 0) = 209.35 KJ

. . Total heat absorbed by ice ‘Q’ from atmosphere

= (41.86 + 666.6 + 209.35) KJ = 917.81 KJ

. Entropy change of the atmosphere

= (ΔS) atm = – Q / T = -917.81 / 273 + 25 = – 3.08 KJ / K

Entropy change of the system (ice) as it gets heated from- 10°C to 0°C

= 0.523 KJ/K

Similarly

(ΔSII) system = 333.3 x 2 / 273 = 1.22 x 2 kj / k

= 2.44 kj/k

= 2 x 4187 in 298 / 273

= 0.733 kj / k

Total entropy change of ice

as it melts to water at 25°C

= (0.523 + 2.44 + 0.733)

= 3.696KJ/K

(ΔS)univ = (ΔS)system + (ΔS)atm

= 3.696 + (- 3.08)

= 0.616KJ/K Ans.

Example 8.4. What is the minimum amount of work necessary to convert one kg of water at 20°C into ice at – 5°C? Take c P of ice = 2.093 KJ/kg k, Latent heat of fusion of ice = 333.3 KJ/kg, C water = 4.187 K]/kg K.

Solution. Heat removal from water at 20°C to convert it to ice at- soc

= 427.5 KJ (student to calculate & match)

Here                (ΔS)ref = 0

( ΔS)atm = Q +W / T = 427.5 + W / 293

( ΔS)system = – 1.5549 kj/k (student to calculate & match)

( ΔS)univ = ( ΔS)sys + ( ΔS)ref +( ΔS)atm

= (S1 – S4) + 0 + 427.5 + W / 293

= – 1.5549 + 427.5 + W / 293 > 0

or                                  427 + W / 293 > 1.5549

or                                 427.5 + W> 455.58

or                                 W > 28.08,Wmin = 28.08 kj.

Example 8.5. How much does the entropy of 1 kg of ice change as it melts into water in each

case (i) Heat is supplied reversibly to mixture of ice and water at 0°C (ii) a mixture of ice and

water at 0°C is stirred by a paddle wheel. (Latent heat of fusion of water at 00C= 335 KJ/Kg)

Solution.

(i)  (ΔS)system = Q rev / T = 335 / 273 + 0 = 1.2271 KJ / K.

(ii) (ΔS)system = 1.2271 kj /k

because the irreversible path is replaced by reversible path. As Sis a point function there will be same entropy rise to both cases. Fig. 8.24

Example 8.6. 3 kg of water at 30°C is mixed with 2 kg of water at 80°C adiabatically at n

constants pressure of 1 atm. Find the increase in the entropy of the total mass of water due to the

mixing (c1 , of water = 4.187 K]/kgk)

Solution.

Let                               m1 = 2 kg, T1 = 273 + 80 = 353 K, c1 = 4.187

m2 = 3 kg, T2 = 273 + 30 = 303 K, c2 = 4.187

using equation (8.20) we get

T f = m1 c1 T1 + m2 c2 T2 / m1 c1 + m2 c2

= 2 x4.187 x 353 + 3 x 4.187 x 303 / 2 x4.187 + 3 x 4.187 = 6762.005 / 20.935 = 323k.

From equation (8.21) we get From equation (8.21) we get

( ΔS)univ = m1 c1 in tf / t1 + m2 c2 in tf t2

= 2 x 4.817 in 323 / 353 + 3 x 4.187 in 323 / 303

= – 0.744 + 0.803 = 0.059 kj / k.