Termo

7-26

Entropy Change of Incompressible Substances 7-52C No, because entropy is not a conserved property.

7-53 A hot copper block is dropped into water in an insulated tank. The final equilibrium temperature of the tank and the total entropy change are to be determined. Assumptions 1 Both the water and the copper block are incompressible substances with constant specific heats at room temperature. 2 The system is stationary and thus the kinetic and potential energies are negligible. 3 The tank is well-insulated and thus there is no heat transfer. Properties The density and specific heat of water at 25°C are ρ = 997 kg/m3 and cp = 4.18 kJ/kg.°C. The specific heat of copper at 27°C is cp = 0.386 kJ/kg.°C (Table A-3). Analysis We take the entire contents of the tank, water + copper block, as the system. This is a closed system since no mass crosses the system boundary during the process. The energy balance for this system can be expressed as
Net energy transfer by heat, work, and mass

Ein − Eout 1 24 4 3

= 0 = ∆U

Change in internal, kinetic, potential, etc. energies

∆Esystem 123 4 4

WATER Copper 50 kg 120 L

or,
∆U Cu + ∆U water = 0 [mc(T2 − T1 )]Cu + [mc(T2 − T1 )]water = 0

where mwater = ρV = (997 kg/m3 )(0.120 m3 ) = 119.6 kg

Using specific heat values for copper and liquid water at room temperature and substituting,
(50 kg)(0.386 kJ/kg ⋅ °C)(T2 − 80)°C + (119.6 kg)(4.18 kJ/kg ⋅ °C)(T2 − 25)°C = 0

T2 = 27.0°C The entropy generated during this process is determined from
T   300.0 K  ∆Scopper = mcavg ln 2  = (50 kg )(0.386 kJ/kg ⋅ K ) ln  353 K  = −3.140 kJ/K  T     1 T   300.0 K  ∆S water = mcavg ln 2  = (119.6 kg )(4.18 kJ/kg ⋅ K ) ln  298 K  = 3.344 kJ/K  T     1

Thus,
∆S total = ∆Scopper + ∆S water = −3.140 + 3.344 = 0.204 kJ/K

7-27

7-54 A hot iron block is dropped into water in an insulated tank. The total entropy change during this process is to be determined. Assumptions 1 Both the