Question

Level II: Thermodynamics (Sketch, reasoning, numerical solution, and assess for credit). A metal smith starts with mass miron of molten iron at T. How much mass of water meter would he need to make the temperature of lead and water 70°C? Set up the problem, but do not solve it. [Use the values in the table below to help you). TABLE 12.4 Specific heats of solids and liquids TABLE 12.8 Melting and boiling temperatures and heats of transformation at standard atmospheric pressure Substance cJ/kg .K) Substance T. (C) LJ/kg) T. (C) Solids L. (J/kg) Lead 128 Nitrogen (N) -210 0.26 x 10 -196 1.99 x 10 Gold 129 Ethyl alcohol 1.09 x 10 78 8.79 x 10 385 Copper Mercury -39 0.11 x 10 357 2.96 x 10 Water 449 3.33 x 10 Iron 22.6 x 10 Lead 900 0.25 x 10 Aluminum 1750 8.58 x 10 Waterice 2090 Mammalian body 3400 100 328 Liquids Mercury Ethyl alcohol Water 140 2400 4190

Answer 1

Guven data - The temperature of lead and Water = 70°c The final temparatwe is 78c and the loud at sasa istoc and the lowd at 328°c by the Given data The Amount or quantity of Water would be added to bring leolds temperatuve to 70 c is as followed Isol Too desected - Too absorted - SQ = sa -M C To S dt - Moto 328 hexe :- Me = Mass of Leaud Mw: Mass of water + MLCL -MWCW So, Mauss of water M = Mouss of lead CLX (308) K CW X (70)k

1. C. - 12 8 3/iy - K- Cw = 4190 7/Kg.K : 1. - MC 29 3/x+-+) (308) (q196 3/%9-4) (1) - M, ( 12%)(306) (u190) (10) - M ( 39,921) 9,93, 300 : ML [o - 1341) ML - Co. 13xy M,)