University:
Blinn College
Course:
PHYS 2325 | University Physics I
Academic year:

2023
Views:

268

Pages:

2
Author:

seburuk5q2

Chapter O - Problems Problem O.1 What is the change in entropy of 3 kg of water initially at 20 °C when it is cooled and frozen resulting in ice at 0°C? Solution to O.1 This is a two step problem. D S1 is the change in entropy when the water is brought from 20 °C to 0 °C and D S2 is the change when the water at 0 °C freezes to 0 °C. When water freezes to ice heat is removed so Q = -m L f . The sign of the entropy of the water to water case follows from the values of the temperatures. The temperature must be in Kelvin in all cases here. Tf D S = D S1 + D S2 = m c ln = 3 µ 4186 ln 273 293 + -3µ3.33µ10 Ti 5 273 + Q T = m cwater ln Tf Ti + - m Lf T = -887.9 - 3659.3 = -4550 J ê K Problem O.2 What is the total change in entropy when 3000 J of heat flows into a house at 20 °C from the warm exterior at 40 °C? Solution to O.2 The entropy of the exterior decreases, since heat flows out of it, and the entropy of the interior increases since heat flows into it. D Stot = D Soutside + D Sinside = -Q Toutside + Q Tinside = -3000 313 + 3000 293 = -9.585 + 10.239 = 0.654 J ê K The second law of thermodynamics says that the total change in entropy cannot be negative for any thermally isolated system. A system is thermally isolated if no heat flow into or out of it. Problem O.3 What is the total change in entropy when a 20 g ice cube at 0 °C is dropped into the Gulf of Mexico, which is at a temperature 15 °C? It is quite safe to assume that the temperature change of the Gulf is negligible. Solution to O.3 The heat that flows into the ice cube comes from the Gulf. This heat Q first melts the ice and then raises its temperature from 0 °C to 15 °C. Q = m L f + m c D T = 0.02 µ 3.33 µ 105 + 0.02 µ 4186 µ 15 = 6660.00 + 1255.80 = 7915.80 Since this heat flows out of the Gulf, which is at constant temperature, its entrop change is D SGulf = -Q T = -Q 288 = -27.485 J ê K The entropy change of the ice involves two terms, the first to melt it and the second to raise its temperature. D Sice = D S1 + D S2 = = 0.02µ3.33µ105 273 + 0.02 µ 4186 ln 288 273 +m L f T + m c ln Tf Ti = 24.396 + 4.478 = 28.874 J ê K The total change is the sum of the two. D Stot = D SGulf + D Sice = 1.39 J ê K 2 Chapter N - Problems Problem O.4 A heat engine absorbs 6000 J of heat from a hot reservoir and expels 4100 J to a cold reservoir at 30 °C. (a) What is the efficiency of this engine? (b) What is the work done by the engine? (c) What is the smallest temperature the hot reservoir could have? Solution to O.4 (a) e = 1 - Qc QH =1- 4100 6000 = 0.31667 = 31.7 % (b) QH = QC + W ï W = QH - QC = 6000 - 4100 = 1900 J (c) The efficiency has an upper limit of emax = 1 - TC TH e § emax = 1 - where T must be in Kelvin. TC TH ï TH = TC 1-e = 303 1-0.31667 = 443.4 K = 170.4 °C Problem O.5 A Carnot engine that operates between a 500 K hot reservoir and a 300 K cold reservoir has a power output of 6000 W. What is the efficiency of this heat engine? At what rate does this engine absorb heat? Solution to O.5 A Carnot engine is a theoretical heat engine of maximum efficiency. e = ec = 1 - TC TH =1- 300 500 = 0.40 = 40 % The power output is the work per time. e= W QH = Wêt QH êt = P QH êt ï Qh ê t = P e = 6000 0.40 = 15 000 W