Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.
1 answer:
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Answer:
The net amount of energy change of the air in the room during a 10-min period is 120 KJ.
Explanation:
Given that
Heat loss from room (Q)= 60 KJ/min
Work supplied to the room(W) = 1.2 KW = 1.2 KJ/s
We know that 1 W = 1 J/s
Sign convention for heat and work:
1. If heat is added to the system then it is taken as positive and if heat is rejected from the system then it is taken as negative.
2. If work is done by the system then it is taken as positive and if work is done on the system then it is taken as negative.
So
Q = -60 KJ/min
In 10 min Q = -600 KJ
W = -1.2 KJ/s
We know that
1 min = 60 s
10 min = 600 s
So W = -1.2 x 600 KJ
W = -720 KJ
WE know that ,first law of thermodynamics
Q = W + ΔU
-600 = - 720 + ΔU
ΔU = 120 KJ
The net amount of energy change of the air in the room during a 10-min period is 120 KJ.
Answer:
(a) The heat transferred is 2552.64 kJ
(b) The entropy change of the mixture is 1066.0279 J/K
Explanation:
Here we have
Molar mass of H₂ = 2.01588 g/mol
Molar mass of N₂ = 28.0134 g/mol
Number of moles of H₂ = 500/2.01588 = 248 moles
Number of moles of N₂ = 1200/28.0134 = 42.8 moles
P·V = n·R·T
V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³
Since the volume is doubled then
V₂ = 2 × 7.25 = 14.51 m³
At constant pressure, the temperature is doubled, therefore
T₂ = 600 K
If we assume constant specific heat at the average temperature, we have
Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂
cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K
cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K
Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 = 2552.64 kJ
b)
Where:
and
are the mole fractions of Hydrogen and nitrogen respectively.
Therefore, = 248 /(248 + 42.8) = 0.83
= 42.8/(248 + 42.8) = 0.1472
∴ = 1066.0279 J/K
Answer:
The answer is "".
Explanation:
Please find the correct question in the attachment file.
using formula: