Conductivity, tough question?

In an aluminum pot, .2 kg at 100 Degrees c boils away in 5 min. The bottom of the pot is 2.5x10^-3 m thick and has a surface area of .02 m^2.to prevent the water from boing too rapidly , a stainless steel plate has been placed between the pot and the heating element.The plate is 1.2x10^-3 m thick, and its area matches that of the pot. Assuming that heat is conducted into the water only through the bottom of the pot , find the tempeture of (a) the aluminum-steel interface and (b) the steel surface in contact with the heating element.





Thermal conductivity:


steel = 14


aluminum = 240





and just incase water = .6





If you could please explain the solution. thank you very much.

Conductivity, tough question?
Water is already at 100 °C. Heat added is used up in changing water from the liquid to the vapor phase. This takes 540 kcal per kg of water, or 2.26 E3 kJ/kg. To boil 0.2 kg of water, 0.2 kg × 2.26 E3 kJ/kg = 452 kJ are required. This is done in 300 s (5 min), so to boil 0.2 kg of water in this time, heat should be delivered at the rate of P = E/t = 452 E3 / 300 = 1 506 J/s, or 1 506 W. This power P customarily appears as Q/t in heat transfer formulas.





a) Now, since P = Q/t = kA ∆T/∆x, solving for ∆T yields





∆T = P ∆x / kA = 1.5 E3 × 2.5 E-3 / 240 × 2 E-2 = 0.785 K = 0.785 °C.





This is the temperature difference between both sides of the aluminum bottom. Since the water is at 100 °C, temperature at the aluminum/steel interface is 100.785 °C.





b) Repeat the calculation above with k = 14 W/m·K, instead of 240. This results in a temperature difference of 6.43 K, or 6.43 °C. Since the colder side of the steel plate is at 100.785 °C, the hotter side will be at 100.785 + 6.43 = 107.2 °C.
Reply:0.2 kg water at 100 C boils away in 5 mins.





The heat required to boil the water away is 0.2*2260000 = 452,000 J.





If this is delivered over 5 mins, then the rate of heat transfer is dQ/dt = 452000/300 = 1507 J/s.





This is transferred across the bottom of the aluminium pot. Using the equation for thermal conductivity





dQ/dt = k*A*dT/dl





1507 = 240*0.02*dT/(0.0025)





Solving gives dT = 0.8 C





One side of the pot is at 100 C (water) so the other side must be at 100.8 C





Now transfer our attention to the steel plate.





We will assume that the same rate of heat transfer exists across this plate. Apply the same equation





dQ/dt = k*A*dT/dl





1507 = 14*0.02*dT/(0.0012)





Solving gives dT = 6.4 C





One side of the plate is at 100.8 C, hence the other side is at 107.2 C