Question

For 1 kg of liquid water, determine the entropy change of the universe when the water is: Initially at 0 C, is heated to 50 C by contact with a thermal reservoir at 50 C and then to 100 C with a thermal reservoir at 100 C Assume Cp = 4.2 kJ/kg-K

Answer 1

Ans

For water

Step 1

Water is heated from 273 K to 323 K

Entropy change of water

S1 = m x Cp x ln (T2/T1)

= 1 kg x 4.2 kJ/kg-K x ln (50+273)/(0+273)

= 0.706 kJ/K

Step 2

Water is heated from 323 K to 373 K

Entropy change of water

S2 = m x Cp x ln (T2/T1)

= 1 kg x 4.2 kJ/kg-K x ln (373)/(323)

= 0.604 kJ/K

Total entropy change of water

Sw = S1 + S2 = 0.706 + 0.604 = 1.31 kJ/K

For reservoir

Step 1

Heat released by thermal reservoir

Q1 = - mass x Cp x (T2-T1)

= - 1 kg x 4.2 kJ/kg-K x (323 - 273)K

= - 210 kJ

Entropy change of reservoir

S1r = Q1/Tr1 = - 210/323

S1r = - 0.650 kJ/K

Step 2

Heat released by thermal reservoir

Q2 = - mass x Cp x (T2-T1)

= - 1 kg x 4.2 kJ/kg-K x (373 - 323)K

= - 210 kJ

Entropy change of reservoir

S2r = Q2/Tr2 = - 210/373

S2r = - 0.563 kJ/K

Total entropy change of reservoir

Sr = S1r + S2r

Sr = - 0.650 - 0.563

Sr = - 1.213 kJ/K

Entropy change of universe

S = Sw + Sr

= 1.31 - 1.213

= 0.097 kJ/K