Question.3757 - Homework Assignment for Exam 2 PHY 100- Fall 2024 Chapter 10 Problems 2) C = 5/9? (F?32) C = 5/9 (20-32) C = - (60/9) = - 6.67 ?C 6) m = 90g Tf =? 88?C Ti = 12?C (c) specific heat of water = 1 cal/g??C ?T = Tf??Ti? = 88?C ? 12?C = 76?C Q = mc (Tf??Ti?) = 90 x 1 (76) = 6840 cal7) m = 300g Tf =? 40?C Ti = 160?C (c) specific heat of water = 0.215 cal/g??C ?T = Tf??Ti? = 40?C ? 160?C = -120?C Q = mc (Tf??Ti?) = 300 x 1 (-120) = -7740 cal8) m = 120 g Lf? = 80 cal/g Q = mLf = 120 g ? 80 cal/g = 9600 cal? 11) 1 cal = 4.184 J Q = 300 cal ? 4.184 J/cal = 1255.2 J 13) a. Work done = 825 J, due to the work done by gas on surrounding it decreases internal energy b. Heat added = 1200 J, causing an?increase in internal energy. c. Net change in internal energy ?U = Q W = 1200 J 825 J = 375 J 14) Vi? = 0.4 m3 Vf = 2.5?m3 P = 3600 pa ?V = Vf ?? Vi ?= 2.5 m3 ? 0.4 m3 = 2.1 m3 W = P?V = 3600 pa x 2.1 m3 = 7560 J work done by gas 17) a. 400 cal of heat added, in converting to joules Q = 400 cal ? 4.184 J/cal = 1673.6 J b. Work done = 400 J, ?U (change in internal energy) = Q W = 1673.6 J 400 J = 1273.6 J c. converting to calories, ?U = 1273.6 J/4.184 J/cal ?? 304.5 cald. The gas will decrease overall temperature because heat is added and work is done on the surroundings. SP3) a. m = 140 g cice = 0.5 cal/g??C Q=mc?T = 140 x 0.5 x (0-(22)) = 1540 cal (Heat required to raise ice to 0?C is 1540 cal) b. Q = mLf ?= 140 g ? 80 cal/g =11200 calc. Q=mc?T = 140 x 1 x (27-0) = 3780 cald. Qtotal ?= Qa? + Qb ?+ Qc ?= 1540 cal + 11200 cal + 3780 cal = 16020 cale. melting the ice takes the most energy due to the high latent heat of fusion compared to the heat required for raising the temperature. f. total heat is found by calculating the latent heat of fusion and adding the heat required to raise the temperature of the water because that method accounts for all energy transfers involved in the process. SP5) Vi ?= 0.17 m3 Ti? = 220 K Tf ?= 1100 K a. Considering Charles law, (Vi/Ti) = (Vf/Tf) Vf = Vi x (Tf/Ti) = 0.17 x (1100/220) = 0.85 m3 b. ?V = Vf? ? Vi ?= 0.85 m3 ? 0.17 m3 = 0.68 m3 c. W = P?V = 1800 Pa ? 0.68 m3 = 1224 J d. Vi ?= 0.24 m3 Vf? = 0.24 x 1100/220 = 1.2 m3 ?V = Vf? ? Vi ?= 1.2 m3 ? 0.24 m3 = 0.96 m3 W = P?V = 1800 Pa ? 0.96 m3 = 1728 J e. Yes, the same amount of gas is not involved because different initial volumes affect the work done in expansion.
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