Question

CHE 120 Pb.3. A 295- aluminum engine part at an initial temperature of 13.00'C absorbs 75.0 kJ of heat. What is the final temperature of the part? (Specific heat capacity of Al = 0.900 Jig K). Pb.4. When 155 mL of water at 26°C is mixed with 75 mL of water at 85°C, what is the final temperature? (Assume no heat lost to the surroundings; density of water = 1.00 g/mL and specific heat capacity of water = 4.184 J/g.K). Pb.5. One piece of copper jewelry at 105°C has twice the mas of another piece at 45°C,. Both are placed in a calorimeter of negligible heat capacity. What is the final temperature inside the calorimet (specific heat capacity of copper = 0.3874 J/g.K)

Answer 1

mass, m = 2959 specific heat capacity, C = 0.900 J/qk Heat absorbed, q = 750kJ = 75.0*109 J Initial temperature, Ti = 13.00°C = (13.00 +273:15) K = 286.15K Let final temperature in K be I. Now q = mcat q=mc (- T:) 75.00 x 10°y = 2959 * 0.900,8 x (7– 286115) 75.60 xlo'- 295x0.900x ( 5p -28615) 75.00x10'= 265.5 I, - 75972-825 75000 + 75972.825 = 265:51 150972.825 = 265.55 150972825 265.5 Thus, final 586.64 temperature in K = 586-64 K = (58664 - 273'15) °C - 1295.5°C] Css Scanne 295.5°C Cams

4. Mass of cold water = 155 ml x 100g/ml 1559 Mass of hot water = 75 m2 x 100g/ml =759 As heat loss by hot water is equal and opposite to heat gain by cold water, Quot = -Q.cold (mea Thut =-(CAT) cold 75.09x&x (1 –855)° X = -155gxgé (-26) 751, - 6375 = -155T;+ 4030 75 T, +155 = 4030 +6375 230 T = 10405 or Is - 10405 a 45.2 Thus, final temperature, It = 45.2°C] Lo 230 Scannet CamScai

5. When m = mass of colder piece then mass of hotter piece = 2m Let Is in oC be the final temperater, Since heat released by hot Copper is equal and opposite to heat absorbed by cold copper, Quot = - Q cold (om CAT) hor = {m CAT) cold 2y4* ¢x (Ts -105)'y = = ykxg* (15 - 45)°4 254-210 = -Ts+45 2T + T = 45+210 3Tg = 255 on T = 255 = 85 Thus, final temperature, Is = [85°C] CS Scann Cams