Answer: a 8143.71 kJ/kg
b 393.15 K
Explanation:
This system is an isobaric process in which there is no change in pressure a quasistatic process where a pressure distribution exists
a since no change in pressure =0 the system does work thus
FOR HELIUM properties in standard thermodynamic chart
cv = 3.1 kJ/kgK
M = Molar mass = 4 kg/kmol
R = Universal gas constant = 8.314 kJ/kg K
cp ≈ cv +R /M = 3.1 + 8.314 /4 = 5.1785 kJ/kgK
Cp = cp * M = 5.1785 kJ/kgK * 4 kg/kmol = 20.714 kJ/kgkmol
T = 120 °C to Kelvin = 120 + 273.15k = 393.15 K
W =n Cp ΔT = 1 kmol * 20.714 kJ/kg kmol* 393.15 K = 8143.71 kJ/kg
b convert T °C = T K thus 120 + 273.15 K = 393.15 K
P₁/T₁ = P₂/T₂
200 kPa/ 393.15 K = 200 kPa/T₂
T₂ = 200 kPa * 393.15 K/ 200 kPa = 393.15 K or 120 k
Answer & Explanation:
function Temprature
NYC=[33 33 18 29 40 55 19 22 32 37 58 54 51 52 45 41 45 39 36 45 33 18 19 19 28 34 44 21 23 30 39];
DEN=[39 48 61 39 14 37 43 38 46 39 55 46 46 39 54 45 52 52 62 45 62 40 25 57 60 57 20 32 50 48 28];
%AVERAGE CALCULATION AND ROUND TO NEAREST INT
avgNYC=round(mean(NYC));
avgDEN=round(mean(DEN));
fprintf('\nThe average temperature for the month of January in New York city is %g (F)',avgNYC);
fprintf('\nThe average temperature for the month of January in Denvar is %g (F)',avgDEN);
%part B
count=1;
NNYC=0;
NDEN=0;
while count<=length(NYC)
if NYC(count)>avgNYC
NNYC=NNYC+1;
end
if DEN(count)>avgDEN
NDEN=NDEN+1;
end
count=count+1;
end
fprintf('\nDuring %g days, the temprature in New York city was above the average',NNYC);
fprintf('\nDuring %g days, the temprature in Denvar was above the average',NDEN);
%part C
count=1;
highDen=0;
while count<=length(NYC)
if NYC(count)>DEN(count)
highDen=highDen+1;
end
count=count+1;
end
fprintf('\nDuring %g days, the temprature in Denver was higher than the temprature in New York city.\n',highDen);
end
%output
check the attachment for additional Information
Answer:
sorry not having idea about answer