Answer to Question #94346 in Molecular Physics | Thermodynamics for Amit Rajaram Shinde

Question

25 kg of water at 95°C mix with 35 kg of water at 35°C, the pressure being taken as constant and temperature of surrounding being 15°C. Total available energy (before mixing) is (Cp of water= 4.2 KJ/kg.k)
a) 1085 KJ
b) 400 KJ
c) 985 KJ
d) 685 KJ

Answer 1

Solution:

The available energy of 25 kg of water at 95°C:

$(A.E.)_{25} = mc_p\int_{T_0}^{T} (1-\frac{T_0}{T})dT =$

$=25\cdot{4.2}\int_{273+15}^{273+95} (1-\frac{288}{T})dT =$

$= 105[(368-288)-288\cdot{ln\frac{368}{288}}] = 987.49$ kJ.

The available energy of 35 kg of water at 35°C:

$(A.E.)_{35} = mc_p\int_{T_0}^{T} (1-\frac{T_0}{T})dT =$

$=35\cdot{4.2}\int_{273+15}^{273+35} (1-\frac{288}{T})dT =$

$= 147[(308-288)-288\cdot{ln\frac{308}{288}}] = 97.59$ kJ.

Total available energy (before mixing) is:

$(A.E.)_{total} = (A.E.)_{25} + (A.E.)_{35} = 987.49+97.59 = 1085$ kJ.

Answer:

Total available energy (before mixing) is 1085 kJ.