Question

Problem 6: [30] An aluminum rod with radius r= 2.0 cm and length L = 18.0 cm is at a temperature of 400 C. It is dropped into an insulating bucket containing 100.0mL of ethanol at 25 C. (a) What are the masses of the rod and ethanol? (see table below) [6] (b) What will be the final temperature of the rod when all the ethanol has vaporized? [15] (c) What is the change in length L of the rod during its cooling? [9] Peth 789 kg/m PAI 2700 kg/m Ceth 2428 J/kg K CAI 910 J/kg K Lethanol 854x10° J/kg Ethanol boil pt 78C αΑΙ 7.21x10-5 /K

Answer 1

Problem 6: an aluminum rod 8=2cm = 0.02m The radies of The length, -0.19 m length, L = 18 cm Temperature, T: 400c (aluminum rods Volume, - 01L = [(*10-") mi Temperature of ethanol, T = 25C V = 100 mL ma * VAL a) The mass of aluminum AL "AL where yolume of aluminum rid aluminum suid Vor A* l Vie - 2 → VAL=T.*.(0.02) * 0.18 = 2,26 2 *10-4 The man, MAL= PA #V VAL z 2700 * 2•262 *10-4 = 0.617 kg ethanol, methanol The mass of ethanol, Pathank & Verhanol = 89 * 1104 b) The heat lon 20.0789 kg by aluminum rod = Heat goured by ethanol

MACAI (400-7) = m e ( 78-2 _35) + mele → 0.611 * 910*(400-T) = 0.0789* 2428 *(53) +(0.0789 *854*103) → 556.01 (400 -T) = 10153.17 +67380.6 400-T = 139.45 T= 260.55 € (②) The change in in length, AL = LAAT AL = 18 * 7.2 * 10°* ( 400 – 260.55) a AL= 0.181 m AL = 18.1 cm