Posted by **Joy** on Saturday, August 24, 2013 at 10:22pm.

A polystyrene ice chest contains 3 kg of ice at 0 degrees Celsius. Triangle H for melting = 3.34 x 10 to the 5 power J/kg p = 1000 kg/m cubed, CP =4190 J/ (kg K). Heat transfer is limited by conduction (k=0.06W/m K)) from the outside wall at 30 degrees Celsius. This means that temperature within the ice chest is not a function of position. Assume the heat flux (J/(sec m squared) is the same through all walls: top, sides, bottom. The walls have thickness = 4 cm and total surface area = 2400 cm squared.

Suppose heat transfer is due to conduction through the wall and convection at the surface with h = 8W/ (m squared K) to ambient air at T infinity = 30 degrees Celsius.

Calculate heat flux (J/ (sec m squared) at the walls.

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