TITLE.PM5

266 ENGINEERING THERMODYNAMICS

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/M-therm/th5-3.pm5

However, at temperatures above absolute zero, the entropy is a function of pressure also.
The absolute entropy of a substance at 1 atm pressure can be calculated using eqn. (5.42) ; for
pressures different from 1 atm, necessary corrections have to be applied.

ENTROPY
Exmaple 5.21. An iron cube at a temperature of 400°C is dropped into an insulated bath
containing 10 kg water at 25°C. The water finally reaches a temperature of 50°C at steady state.
Given that the specific heat of water is equal to 4186 J/kg K. Find the entropy changes for the
iron cube and the water. Is the process reversible? If so why? (GATE, 1996)
Solution. Given : Temperature of iron cube = 400°C = 673 K
Temperature of water = 25°C = 298 K
Mass of water = 10 kg
Temperature of water and cube after equilibrium = 50°C = 323 K
Specific heat of water, cpw = 4186 J/kg K
Entropy changes for the iron cube and the water :
Is the process reversible?
Now, Heat lost by iron cube = Heat gained by water
mi cpi (673 – 323) = mw cpw (323 – 298)
= 10 × 4186 (323 – 298)

∴ mi cpi =

10 4186 323 298

623 323

×−

−

()

() =^2990

where,mi = Mass of iron, kg, and

cpi = Specific heat of iron, J/kg K

Entropy of iron at 673 K = mi cpi ln 673

273

F

HG

I

KJ

= 2990 ln

673

273

F

HG

I

KJ

= 2697.8 J/K [Taking 0°C as datum]

Entropy of water at 298 K = mw cpw ln

298

273

F

HG

I

KJ

= 10 × 4186 ln

298

273

F

HG

I

KJ = 3667.8 J/K

Entropy of iron at 323 K = 2990 × ln

323

273

F

HG

I

KJ = 502.8 J/K

Entropy water at 323 K = 10 × 4186 ln

323

273

F

HG

I

KJ = 7040.04 J/K

Changes in entropy of iron = 502.8 – 2697.8 = – 2195 J/K

Change in entropy of water = 7040.04 – 3667.8 = 3372.24 J/K

Net change in entropy = 3372.24 – 2195 = 1177.24 J/K

Since ∆S > 0 hence the process is irrevesible. (Ans.)

Example 5.22. An ideal gas is heated from temperature T 1 to T 2 by keeping its volume

constant. The gas is expanded back to its initial temperature according to the law pvn = constant.

If the entropy change in the two processes are equal, find the value of n in terms of the adiabatic

index γ. (U.P.S.C., 1997)