Question

FR 5. (15 points) Two blocks one of aluminum (Al MM= 26.98g/mol , Cp,m=24.35 J/mol K), the other of silver (Ag MM= 107.9 g/mol , Cp,m 25.351 J/mol K) of mass (100 g) are at two different temperatures (TAI=300K and Tag=500K). Calculate the change in entropy for the system once the final temperature is achieved.

Answer 1

→ Given data - Mass of Al bli Molar specific to Molar 107.9gtmd - 0.9267 mde of Al block = Mass of Ag block om olar specific heat capacity (com) of 'AN = 24.35 J/molk lan specific heat capacity (Aim) of 'Ag' = 25.351J/mock holar mass of Ag = 107,9 g/mol olar mass of A1 = 26.98 g/mol No of molescm) of Ag - hoog > Nos of molecol of AL = 100g almal = 3.71malo > Initie temperature of All = T, = 3ook s Initial temperature of "Ag = T = 5ook - Assuming final temperature of the system be TE Heat gained by Al block = Heat lost by Ag block conx (F-TY) = n' cpom (T3-T6) 26 8 mo

3. Il mole 24.25J/mol.kx (T-Book) - 0.9264 mee x olok x (TF - Book) = 0.9264 mde X 25.3515/molla x (500-If) [Tf 300 l = 500-TCL 0.26005 DO 341. 29 KN Tf - 300 = 130.02 – 0.26 005TR TE : 341-29K > final 9 Temperature (Of) of system = 341. 29K Total Entropy change ( Assya) = As, +952 = mcpim en Chi) = 3.41mde X 24,35J/mol.k en goo29 As = 11.649 J/K SA = Entropy change of Ag AS, --n! Comen Cup = -0.9264 mole X 25-3519/ 701.k x ln (Santiagok) AS2 - 8.97 g/k Total Entropy change ( 45 sya) for the system - AS, TAS, As sye = 11.649 J/ k 8.97J/k 2.675/K Total Entropy change (Assys) = 2.675k = ASSYB 2062 THE = 2.67 Ilk Toto