Answer:
q_poly = 14.55 KJ/kg
Explanation:
Given:
Initial State:
P_i = 550 KPa
T_i = 400 K
Final State:
T_f = 350 K
Constants:
R = 0.189 KJ/kgK
k = 1.289 = c_p / c_v
n = 1.2 (poly-tropic index)
Find:
Determine the heat transfer per kg in the process.
Solution:
-The heat transfer per kg of poly-tropic process is given by the expression:
q_poly = w_poly*(k - n)/(k-1)
- Evaluate w_poly:
w_poly = R*(T_f - T_i)/(1-n)
w_poly = 0.189*(350 - 400)/(1-1.2)
w_poly = 47.25 KJ/kg
-Hence,
q_poly = 47.25*(1.289 - 1.2)/(1.289-1)
q_poly = 14.55 KJ/kg
its period should be the amount it takes to cycle from cycle to cycle so it would be 10 and your frequency would have to be calculated by taking 10 and dividing by 2 since that is how many cycles you have done so your frequency is 5
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Answer:
Check explanation
Explanation:
Gold - Au (Aurum)
Mercury - Hg (Hydrargyrum)
Copper - Cu (Cuprum)
Iron - Fe (Ferrum)
Lead - Pb (Plumbum)
These elements in the periodic table are some of the elements represented by letters not in line with their names.
This is because, these elements were known in ancient times and therefore, they are represented by letters from their ancient names.
La masa molar de 65 litros de SO2 es igual a 64,1 g/mol.
<h3>Masa molar</h3>
La masa molar de un compuesto depende de su masa presente en 1 mol, entonces:
Para calcular la masa molar de un compuesto, simplemente suma las masas de cada elemento en el compuesto, así:
Así, la masa molar de 65 litros de SO2 es igual a 64,1 g/mol.
Obtenga más información sobre la masa molar en: brainly.com/question/17109809
Answer:
The latent heat of vaporization of water is 2.4 kJ/g
Explanation:
The given readings are;
The first (mass) balance reading (of the water) in grams, m₁ = 581 g
The second (mass) balance reading (of the water) in grams, m₂ = 526 g
The first joulemeter reading in kilojoules (kJ), Q₁ = 195 kJ
The second joulemeter reading in kilojoules (kJ), Q₂ = 327 kJ
The latent heat of vaporization = The heat required to evaporate a given mass water at constant temperature
Based on the measurements, we have;
The latent heat of vaporization = ΔQ/Δm
∴ The latent heat of vaporization of water = (327 kJ - 195 kJ)/(581 g - 526 g) = 2.4 kJ/g
The latent heat of vaporization of water = 2.4 kJ/g
