Please answer all the questions! An engine absorbs 1.70 kJ from a hot reservoir
ID: 2305635 • Letter: P
Question
Please answer all the questions!
An engine absorbs 1.70 kJ from a hot reservoir at 277°C and expels 1.20 kJ to a cold reservoir at 27.0°C in each cycle. What is the net change in internal energy of the heat engine in each cycle? 9. 0.50 kJ 1.20 kJ 1.70 kJ 2.90 kJ 10. What is the entropy change of 0.50 kg of water initially at 25°C that is completely frozen into ice at -10°C? Given that: cice 2,090 J/(kg.K), Lf 3.34 x 105 J/kg, Cwater 4,186 J/(kg.K), and ? 2.256 × 106 J/kg a. -390 J/K b. -830 J/K c. -2,300 J/K d. -100,000 J/K e. -230,000 J/K f. None of the above as the answer is positive 11. For a potential energy curve (i.e., a plot of potential energy U(x) versus position) what are the conditions on the first and second derivatives of the potential energy function for an unstable equilibrium point? a. U'>0 & U"> 0Explanation / Answer
9. (a) When a system is returning to its original state at the end, the internal energy is 0 as the variables trace same path. The engine always returns to the initial state after each cycle, so the change in internal energy is always 0
10. Change in entropy when the matter is in same state = dS = m*c*ln (Tfinal/Tinitial)
m =mass
c= specific heat
Tfinal and T initial are the final and initial temperatures respectively in kelvin
during the conversion of 1 state to another, dS = dQ/T = m*L/T
when water temperature is reduced from 25 to 0 degrees,
dS1 = 0.5*4186*ln(273/298)
= -183.83 J/K
when water is converted to ice at 0 degrees
dS2 = m*Lf /T = 0.5 *334000/273 = -611.72 J/K
when ice temp is reduced from 0 to -10
dS3 = 0.5 *2090*ln(263/273) = -38.99 J/K
total change in entropy = dS1 + dS2 +dS3 = -834.547 J/K
so answer is -830 J/K ( option b)
11. An unstable equilibrium point is the local maxima of the curve
in order to be a maxima, du/dx = 0 and d2u/dx2 <0
hence, u' = 0 and u'' <0 ---- option ( f ) is correct choice