Answer:
<h2>698.3Kpa</h2>
Explanation:
Step one:
given data
V1=0.25m^3
T1=290k
P1=100kPa
V2=0.5m^2
T2=405k
P2=? final pressure
Step two:
The combined gas equation is given as
P1V1/T1=P2V2/T2
Substituting we have
(100*0.25)/290=P2*0.05/405
25/290=0.5P2/405
0.086=0.05P2/405
cross multiply
0.086*405=0.05P2
34.9=0.05P2
divide both sides by 0.05
P2=34.9/0.05
P2=698.3Kpa
<u>Therefore the new pressure is 698.3Kpa when the gas is compressed</u>
BRIGHT HEADLIGHTS
AND SEVERE WEATHER CONDITIONS
Answer:
COP = 0.090
Explanation:
The general formula for COP is:
COP = Desired Output/Required Input
Here,
Desired Output = Heat removed from water while cooling
Desired Output = (Specific Heat of Water)(Mass of Water)(Change in Temperature)/Time
Desired Output = [(4180 J/kg.k)(3.1 kg)(25 - 11)k]/[(12 hr)(3600 sec/hr)]
Desired Output = 4.199 W
And the required input can be given as electrical power:
Required Input = Electrical Power = (Current)(Voltage)
Required Input = (2.9 A)(16 V) = 46.4 W
Therefore:
COP = 4.199 W/46.4 W
<u>COP = 0.090</u>
Answer:
W=2 MW
Explanation:
Given that
COP= 2.5
Heat extracted from 85°C
Qa= 5 MW
Lets heat supplied at 150°C = Qr
The power input to heat pump = W
From first law of thermodynamics
Qr= Qa+ W
We know that COP of heat pump given as



W=2 MW
For Carnot heat pump


2.5 T₂ - 895= T₂
T₂=596.66 K
T₂=323.6 °C