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.
- Dr. ZorroLv 74 years ago
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
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.
- lazlo_deiLv 54 years ago