# Help with thermodynamics: entropy?

1. A mass of 1 kg of water at a temperature of 25 ° C

Adiabatically and isobarically with a mass of 3 kg of water at 75 ° C. Calculate

The entropy change of water and the universe. Take cp = 1 cal / g ° K for water.

Relevance
• dS = dQ / T = Cp dT / T

= m cp dT / T

ΔS = m cp ln(Tf/Ti)

The final temperature after mixing is

Tf = (m1 T1 + m2 T2)/(m1+m2) = 62.5 Celsius

For the entropy calculations you need absolute temperatures:

Tf = (273.15 + 62.5) K = 335.65 K

The entropy change of the initially cold mass of water therefore is

ΔS = 1kg * 1 kcal/(kg K) * ln(335.65/298.15)

= 0.11847 kcal/K

The change in entropy of the initially hot water is

ΔS = 3kg * 1 kcal/(kg K) * ln(335.65/348.15)

= - 0.10969 kcal/K

The entropy change of the total mass of water therefore is

8.78 cal/K

The entropy of the environment does not change (no heat is exchanged with it, as the process is said to be adiabatic).

Therefore the entropy change of the universe (system + surroundings) is 8.78 cal/K.