The rate of gain for the high reservoir would be 780 kj/s.
A. η = 35%
W =
W = 420 kj/s
Q2 = Q1-W
= 1200-420
= 780 kJ/S
<h3>What is the workdone by this engine?</h3>B. W = 420 kj/s
= 420x1000 w
= 4.2x10⁵W
The work done is 4.2x10⁵W
c. 780/308 - 1200/1000
= 2.532 - 1.2
= 1.332kj
The total enthropy gain is 1.332kj
D. Q1 = 1200
T1 = 1000
= 1200 - 369.6/1200
= 69.2 percent
change in s is zero for the reversible heat engine.
Read more on enthropy here: brainly.com/question/6364271
Answer: the theoretical density is 4.1109 g/cm³
Explanation:
first the image of one set of ZnS bonding in the crystal structure, we calculate the value of angle θ
θ + ∅ + 90° = 180°
θ = 90° - ∅
θ = 90° - ( 109.5° / 2 )
θ = 35.25°
next we calculate the value of x from the geometry
given that; distance angle d = 0.234
x = dsinθ
= 0.234 × sin35.25°)
= 0.135 nm = 0.135 × 10⁻⁷ cm
next we calculate the length of the unit cell
a = 4x
a = 4(0.135)
a = 0.54 nm = 0.54 × 10⁻⁷ cm
next we calculate number of formula units
n' = (no of corner atoms in unit ell × contribution of each corner atom in unit cell) + ( no of face center atom in a unit cell × contribution of each face center atom in a unit cell)
n' = 8 × 1/8) + ( 6 × 1/2)
= 1 + 3
= 4
next we calculate the theoretical density using this equation
P = [n'∑(Ac + AA)] / [Vc.NA]
= [n'∑(Ac + AA)] / [(a)³NA]
where the ∑Ac is sum of atomic weights of all cations in the formula unit( 65.41 g/mol)
∑AA is the sum of weights of all anions in the formula unit( 32.06 g/mol)
Na is the Avogadro’s number( 6.023 × 10²³ units/mole)
so we substitute
P = [4( 65.41 + 32.06)] / [ ( 0.54 × 10⁻⁷ )³ × (6.023 × 10²³)]
= 389.88 / 94.84
= 4.1109 g/cm³
therefore the theoretical density is 4.1109 g/cm³
