A pot with a steel bottom 8.50 mm thick rests on a hot stove. The area of the bottom of the pot is 0.150 m2. The water inside the pot is at 100.0o C, and 0.390 kg are evaporated every 3.00 min. Find the temperature of the lower surface of the pot, which is in contact with the stove.

Solution 69E Given m=0.390 k 3 Lv=2256 × 10 J/kg 2 A=0.150 m . 0 T c 100 C T H = We know that Q = mL v 3 = 0.390 kg × 2256 × 10 J/kg = 8.798 × 10 J Q H = t 5 = 8.798×10 180 s 3 = 4.888 × 10 J/s We have equation H = kA(THTC) L HL T H...