An ice-making machine inside a refrigerator operates in a Carnot cycle. It takes heat from liquid water at 0.0 ∘C and rejects he
1 answer:
Answer:
2.36 x 10^6 J
Explanation:
Tc = 0°C = 273 K
TH = 22.5°C = 295.5 K
Qc = heat used to melt the ice
mass of ice, m = 85.7 Kg
Latent heat of fusion, L = 3.34 x 10^5 J/kg
Let Energy supplied is E which is equal to the work done
Qc = m x L = 85.7 x 3.34 x 10^5 = 286.24 x 10^5 J
Use the Carnot's equation
QH = 309.8 x 10^5 J
W = QH - Qc
W = (309.8 - 286.24) x 10^5
W = 23.56 x 10^5 J
W = 2.36 x 10^6 J
Thus, the energy supplied is 2.36 x 10^6 J.
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<u>Answer :</u>
(a) The Initial temperature of the steam is 247.557°C
(b) The enthalpy change per unit mass of the steam is -1771 kJ/kg
(c) The final pressure is 1555 kPa, and the liquid vapor mixture contains small mass of vapor.
<u>Explanation :</u>
From the given statements, we know
Initial steam pressure () = 3.5 MPa
Degree of Super heat = 5°C
Saturation Temperature () = 242.557°C [From Steam Table]
Now, the Initial temperature () of steam is given by-
=
+ Degree of Superheat
= 242.557 + 5 = 247.557°C
(b)To find: The enthalpy change per unit mass of the steam is
We know enthalpy (h) change is given by
Δh = –
From steam Table, properties of the gas at state 1 and state 2
= 2821.1 kJ/kg
= 1049.7 kJ/kg
On substituting the values, we get
or, Δh = 1049.7 - 2821.1 = -1771 kJ/kg
(c) To Find: The final pressure of the liquid vapor mixture contains small mass of vapor.
We find that the gas at the final position, i.e., at position 3
Final Temperature () = 200°C [Given] ……………………………..(1)
Specific Volume () = Specific Volume (
) = 0.00123
/kg ….....(2)
Using steam table for corresponding values of (1) and (2), we get
Final Pressure = 1555 kPa
Dryness fraction () = 0.0006.
Answer:
=(5.7 x
i - 2.24 x
j) kgm/s
Explanation:
Due to earths gravity, force on asteroid is given by:
r^
Plugging in the values, we have
F= [(6.67x)(1500)(5.97 x
)(8x
i + 9x
j)] / ((8x
)² + (9x
)²
F= 2736 i^ + 3078 j^
In order find the final momentum of the Asteroid, apply impulse momentum theorem
=
+ FΔt
= 1500(3.5 x
i - 1.8x
j) + (2736i + 3078j)(1.5x
)
=(5.7 x
i- 2.24 x
j)kgm/s