A pot with a steel bottom 8.25 mm thick rests on a hot stove. The area of the bo
ID: 2243082 • Letter: A
Question
A pot with a steel bottom 8.25mmthick rests on a hot stove. The area of the bottom of the pot is 0.175m2. The water inside the pot is at 100?C, and 0.370kgare evaporated every 3.25min. Find the temperature of the lower surface of the pot, which is in contact with the stove.
I know there are already a couple questions on here like this, but I just can't seem to get the right answer when I follow the calculations. Please help me out if you can!
A pot with a steel bottom 8.25mmthick rests on a hot stove. The area of the bottom of the pot is 0.175m2. The water inside the pot is at 100?C, and 0.370kgare evaporated every 3.25min. Find the temperature of the lower surface of the pot, which is in contact with the stove.Explanation / Answer
Q=m*Lv =0.37*(2256*10^3) =834,720 J
H=Q/t =834,720/(3.25*60)
H=4280.6 J/s
H=KA*dT/L
dT =HL/KA
dT =4280.6*(8.25*10^-3)/(50.2*0.175)
dT =4.02 C
dT =Th-Tl
Th =Tl+4.02=100+4.02
Th=104.02 C
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.