1a) A polystyrene ice chest contains 3 kg of ice at 0 C. Delta H for meting= 3.3

Question

1a)                                 A polystyrene ice chest contains 3 kg of ice at 0 C. Delta H for meting= 3.34E5 J/kg, p=1000kg/m^3, Cp= 4190J/(kg*K).                             

                                Heat transfer is limited by conduction (k=0.06W/(m*K)) from the outside wall at 30 C. This means that temperature within the ice chest is not a                                 function of positon.                             

                                Assume the heat flux(J/(sec*m^2) is the same through all walls:top sides, bottom.                             

                                The walls have thickness= 4cm and total surface are= 2400cm^2

question) suppose heat transfer in problem (1a) is due to conduction through the wall and convection at the surface with h=8W/(m^2*K) to ambient air at T ?= 30 C.

a) calculate the outside surface temperature

      *I think I have to use q=hA (Ts-T ?) this equation.

b) calculate heat flux( J/(sec*m^2) at the walls.

Explanation / Answer

q = k A dT / s (1)

where

A = heat transfer area (m2, ft2)

k = thermal conductivity of the material (W/m.K or W/m oC, Btu/(hr oF ft2/ft))

dT = temperature difference across the material (K or oC, oF)

s = material thickness (m, ft)

then

B)heat flux=0.06*2400*10^-4*30/.004=108kw=108J/(sec*m^2

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