Question

Problem 8: Two amounts of salt water at different temperatures are brought into thermal contact with each other and mixed thoroughly. The first amount of water has mass m, = 0.302 kg and its initial temperature is T, = 7.61 °C. The second has a mass of m2 = 0.873 kg and an initial temperature of T2 = 34.5 °C. Assume that the system is isolated. Let the specific heat of the water at this salinity to be Cw=0.84 kcal / kg /K. Part (a) After both amounts of water are mixed and reach thermal equilibrium, what is the total change in the amount of internal energy, in joules, of the system? Numeric : A numeric value is expected and not an expression. AU = - Part (b) Find an expression for the final temperature of the water after thermal equilibrium has been reached. Expression: Ts= Select from the variables below to write your expression. Note that all variables may not be required. a, b, 0, d, g, h, j, k, m,m,m2, P, t, T1, T2 Part (c) Find the change in entropy of the first amount of water alone as it moves from its initial temperature T, to its final temperature Trin kcal/K. Numeric : A numeric value is expected and not an expression. 4S7 = Part (d) Find the total change in entropy of the system, in kcal/K. Numeric : A numeric value is expected and not an expression. AS =

Answer 1

m = 0.302 kg Ti - 7.610 m2 - 0.873 kg Tz = 34.5°C Cw = 0.84KG Tz Applying ist law of thermodynamics - AQ = AutoWt where, sa = change in heat DW = work done AQ = 0 (o system is su= change in internal isolated. on=0 (No work) from i 0 = Auto energy. Total change is internal energy. = 0 b) change in internat energy of t & change in internal enery of 2 = 0 Ulft Upf = 0 m, cw (Tf -T.) + M₂ CW ( T t - 12) = 0 m, Tf - myT, + m₂ Tf - m₂ Tz (m, tme) Tf = m, T + m₂ Tz 1Tf = m + m₂ Tz mitm. expression Ti 0.302 X 7.61 +0.813X 345 0.302 +0.813 If = 27.58°C]

C change in entropy of mass m, is given by - AS, = m, cw en II TJ = 300.58k T, 2 280.61k = 0.302 x 0.84 ln ( 300.58 ) T2 = 301.5 - 0.0175 kcal/k. change in entropy of tofal System. AS = DS + BS2 280.61 d) = my wen If + my Cow les If T T2 300.58 = 0.802 80.84 en 300°SS ) + 0.873.X0.84 les ca 307.5 0.0174400 - 0.016691 BS = 0.0007490 kcallk