H,0 0.2 A m = 25 g P= P, = 300 kPa 120 V Sat. vapor 5 min O=3.7 kJ P. kPa A 300 2.
H,0 0.2 A m = 25 g P= P, = 300 kPa 120 V Sat. vapor 5 min O=3.7 kJ P. kPa A 300 2.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Question
A piston–cylinder device contains 25 g of saturated water vapor that is maintained at a constant pressure of 300 kPa. A resistance heater within the cylinder is turned on and passes a current of 0.2 A for 5 min from a 120-V source. At the same time, a heat loss of 3.7 kJ occurs. (a) Show that for a closed system the boundary work Wb and the change in internal energy ΔU in the first-law relation can be combined into one term, ΔH, for a constant-pressure process. (b) Determine the final temperature of the steam.
![H,0
0.2 A
m = 25 g
P= P, = 300 kPa
120 V
Sat. vapor
5 min
O=3.7 kJ
P. kPa A
300
2.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3e3c7697-88bc-4bb3-be07-4a1e2327abc5%2Fd55e07c9-4bce-40aa-870e-5a416e1f3b45%2Fen5z1s9.png&w=3840&q=75)
Transcribed Image Text:H,0
0.2 A
m = 25 g
P= P, = 300 kPa
120 V
Sat. vapor
5 min
O=3.7 kJ
P. kPa A
300
2.
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