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Please show your work step by step and make sure the answer are correct please!! because I cannot answer it twice
Exercise 17.63-Enhanced-with Solution Constants Part A A pot with a steel bottom 7.00 mm thick rests on a hot stove. The area of the bottom of the pot is 0.200 m The water inside the pot is at 100.0 °C, and 0.370 kg are evaporated every 3.00 min Find the temperature of the lower surface of the pot, which is in contact with the stove. You may want to review (Pages 565-572) For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Conduction through two bars in contact Submit Request Answer Provide Feedback
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Answer #1

Latent heat of vaporization of water is L v = 2256 × 103 J/kg

Q = mL,-(0.370 kg) (2256 × 103 J/kg) = 8.35 × 105 J

H-Q/t = (8.35 × 105 J)/(180 s) 4.638 × 103 J/s

HL (4.638×103 J/s)(7.00×10-3 m) Then H = kA(TH-Tc) / L says that TH-Tc =-= kA(50.2 W/m.K) (0.200m2)

H = 3.23°C

T, = T, 3.23°C = 100°C + 3.23°C-103.23°C 103.2°C

The temperature of the lower surface of the pot, T = 103.2 o C

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