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2.1. You are asked to calculate the electric power required (in kWh) to heat all an alluminium wire (positioned) in a vacuum similar to a light bulb filament from 25C to 660C (liguid) to be used in a vapor deposition apparatus. The melting point of Al is 660C. The wire is 2.5 mm in diameter and has a length of 5.5 cm. (The vapor deposition occurs at temperature in the vicinity of 900C). (10) Data: For AI,Cp =20.0+ 0.0135 T,where T is in K and Cp is in J/(g mol)C.The Husion =10.670 J/(gmol(C) at 660C.The density of AI is 19.35 g/ cm3 2.2. One hundred kg of HCl gas are cooled from 300 to 150C at 1 atm pressure.Calculate H and U in kJ.The heat capacity equation found is: Cp=7.24 1.76 10-3T +3.07 10-6T2 - 10-T3 kg cal Where Cp is in and T is in K (kg mol)(K) (15) 23. What is the enthalpy change that takes place when 3 kg of water at 101.3 kPa and 300K are vaporized to 15 000 kPa and 800 K? (5) [10 Marks] 

          2.1. You are asked to calculate the electric power required (in kWh) to heat all an alluminium
wire (positioned) in a vacuum similar to a light bulb filament from 25C to 660C (liguid) to
be used in a vapor deposition apparatus. The melting point of Al is 660C. The wire is 2.5
mm in diameter and has a length of 5.5 cm. (The vapor deposition occurs at temperature
in the vicinity of 900C).
(10) Data: For AI,Cp =20.0+ 0.0135 T,where T is in K and Cp is in J/(g mol)C.The Husion
=10.670 J/(gmol(C) at 660C.The density of AI is 19.35 g/ cm3
2.2. One hundred kg of HCl gas are cooled from 300 to 150C at 1 atm pressure.Calculate H
and U in kJ.The heat capacity equation found is:
Cp=7.24  1.76 10-3T +3.07  10-6T2 - 10-T3 kg cal Where Cp is in  and T is in K (kg mol)(K)
(15)
23. What is the enthalpy change that takes place when 3 kg of water at 101.3 kPa and 300K
are vaporized to 15 000 kPa and 800 K?
(5) [10 Marks]
Show more…
21 you are asked to calculate the electric power required in kwh to heat all an alluminium wire positioned in a vacuum similar to a light bulb filament from 25c to 660c liguid to be used in 69754

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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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2.1. You are asked to calculate the electric power required (in kWh) to heat all an alluminium wire (positioned) in a vacuum similar to a light bulb filament from 25C to 660C (liguid) to be used in a vapor deposition apparatus. The melting point of Al is 660C. The wire is 2.5 mm in diameter and has a length of 5.5 cm. (The vapor deposition occurs at temperature in the vicinity of 900C). (10) Data: For AI,Cp =20.0+ 0.0135 T,where T is in K and Cp is in J/(g mol)C.The Husion =10.670 J/(gmol(C) at 660C.The density of AI is 19.35 g/ cm3 2.2. One hundred kg of HCl gas are cooled from 300 to 150C at 1 atm pressure.Calculate H and U in kJ.The heat capacity equation found is: Cp=7.24 1.76 10-3T +3.07 10-6T2 - 10-T3 kg cal Where Cp is in and T is in K (kg mol)(K) (15) 23. What is the enthalpy change that takes place when 3 kg of water at 101.3 kPa and 300K are vaporized to 15 000 kPa and 800 K? (5) [10 Marks]
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Transcript

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00:01 Hi, in the first part of the question by heat energy equation we have k into a into delta t divided by l where k is the thermal conductivity, a is the area of cross section, delta t is the temperature, l is the length of the rod.
00:15 So now we have to find area of cross section a which is equal to pi into r square.
00:20 So we have diameter of the rod is 0 .02 meter.
00:25 So radius of the rod will be 0 .01 whole square.
00:29 So into pi.
00:30 So then we have k is equal to, so we have to find the thermal conductivity.
00:35 So k will be equal to q into l.
00:40 So q into l divided by a into delta t.
00:45 So q into l can be written as a into delta x divided by a into delta t.
00:49 So on substituting the value we have 100 watt into delta x is the compression length.
00:55 So or the length of the rod which is 0 .04 meter, area of cross section is pi into 0 .01 whole square into change in temperature is 31 .75.
01:08 So by this calculation we have thermal conductivity is equal to 401 watt per millimeter kelvin.
01:18 Then in the second part of the question we have q is equal to ha into delta t where q is the heat transfer.
01:27 So we have, so we have heater energy is given as it is p into efficiency.
01:34 So we have 1 into 10 power 3 into 0 .95.
01:39 Now we have, so we have to find the delta t...
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