Conductivity and thermodynamics?

In an aluminum pot, .2 kg of water 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.


Additional Details

Conductivity and thermodynamics?
For simple one dimensional heat flow, Fourier's law of conduction takes the form





q = - k · dT/dx





q heat flux


k thermal conductivity


T temperature


x coordinate in direction of heat flow





Assume that all the heat is transferred from the heating element to the water. Than the heat flow through iron plate and the bottom must be constant:


Q = q · A = const


Because the area of the heat transfer A is constant, the heat flux q must constant too.





Therefore


q = - k · dT/dx = const


integrate


∫ dT = - q/k · ∫ dx


with the boundary conditions T(x₀) = T₀


T = T₀ - q/k ·(x - x₀)


which is a linear temperature profile.





The heat flow can be found from the ratio of the vaporising water. The energy transferred per unit time is equal to energy needed to vaporize the given amount of water z :


Q = Δm · ΔHv / Δt


(where ΔHv is the heat of vaporization ΔHv = 2257kJ/kg


thus


q = Q/A = (Δm · ΔHv) / (Δt · A )


= (2kg · 2257kJ/kg) / (300s · 0.02m² ) = 752.3 kW/m²





(a)


At the interface of pot and boiling water the temperatures of both media are the same:


T₀ = 100°C


T = T₀ + q/k ·Δx


= 100°C + 752.3 kW/m² / 240 W/Km² · 0.0025m = 107.8°C





(i change the sign before the second term of the formula because the distance is measured against the direction of q)





(b)


T₀ = 107.8°C


T = T₀ + q/k ·Δx


= 100°C + 752.3 kW/m² / 14 W/Km² · 0.0012m = 172.3°C

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