Question

3. (10 points) Treat a stone of 10 kg and a water of 100 kg as a system. Initially, the stone locates 10 m above the water, and the temperature of the stone and the water are same. Now, the stone falls toward the water. Determine total internal energy change, kinetic energy change, potential energy change, heat transfer, and work for each case with respect to the initial state. Treat g = 9.80665 m/s". a) at the moment just before the stone touch the water b) After the stone falls into the water, the stone becomes stationary. c) The temperature of the stone and the water become same to the initially value.

Answer 1

Answer: Griven that, Toreat a stone of 10 kg and a water of 100 kg as a system. mg = 10 kg mw = 100 kg Stone locale iom above the water h om Temperature of Stone and water au same. 9 = 9.80665 m/s (a) At the moment just before the stone touch the water Considen, potential Energy equation, AEp = mx x oz where m=ms 9c = 1 kg-mins A2ch.-10m -hen, дер — mg x xh Substitute requisud value AEp = 10 x 9180665x610) AEP = -980.665 J. .: potential Energy change is -980.665J some aisl faiction takes place when Stone fall through aig. is negligible. There is no change in temperature l.e no change in internal engigy of o change in internal energy of the system so, AU-0 Internal Energy change =0 J transter takes place due to Temperature of Stone K water same treat transfer -OJ No wölk corosses boundary of system, work = 0 According to ist Jow, P-W - AU+AEK+ AEP AEK = Q-W-AU - AEP

Kinetic energy change = 980.665 J (6) After the Stone -falls into the water, the Stone becomes Slation - Apply 1st law to process from (9) -(3) In State 1, Slone ke =0 (3) (4) Ekini = OJ Ekin2 = 980.6657 when stone hits bottom of tank AEK 192:0 For 2-3 Kinetic energy change =-980.665J According to ist law, AU + AEK-0 AU-980.665 - 0 AU = 980.665 3 Internal Energy change - 980.665 J 023 = Heat transfer=OJ water depth negligible so Potential Energy change -O No wölk Colosses boundary of system w23 = work : 0 (0) The temporature of the stone and water become same to initid value Apply 18t law to porocess from (3) - (4) From ist law, Q34 = AU AU34 = 04-03 = - (03-01) E - AU23 34 = -980.665 J Augu = Internal Energy change = -980.6655 Heat transfer= -980.6653 No wok corosses the boundary wolk - 0 In kinetic and potential energy there is no change thence change in K.E and PE = 0