Frozen food items can be defrosted by simply leaving them on the counter, but it takes too...

Frozen food items can be defrosted by simply leaving them on the counter, but it takes too long. The process can be speeded up consider ably for flat items such as steaks by placing them on a large piece of highly conducting metal, called the defrosting plate ,which serves as a fin. The increased surface area enhances heat transfer and thus reduces the defrosting time.

Consider two 1.5-cm-thick frozen steaks at –18°C that resemble a 15-cm-diameter circular object when placed next to each other. The steaks are now placed on a l-cm-thick black anodized circular aluminum defrosting plate (k = 237 W/m·K, α: = 97.1 × 10-6 m2/s, and s = 0.90) whose outer diameter is 30 cm. The properties of the frozen steaks are p = 970 kg/m3 cp = 1.55 kJ/kg·K, k= 1.40 W/m·K, α = 0.93 X 10-6 m2/s, and ε = 0.95, and the heat of fusion is hif= 187 kJ/kg. The steaks can be considered to be defrosted when their average temperature is 0°C and all of the ice in the steaks is melted. Initially, the defrosting plate is at the room temperature of 28°C, and the wooden countertop it is placed on can be treated as insulation. Also. the surrounding surfaces can be taken to be at the same temperature as the ambient air, and the convection heat transfer coefficient for all ex posed surface scan be taken to be 12 W/m2 · K. Heat transfer from the lateral surfaces of the steaks and the defrosting plate can be neglected. Assuming one-dimensional heat conduction in both the steaks and the defrosting plate and using, the explicit finite difference method, determine how long it will take to defrost the steaks. Use four nodes with a nodal spacing of Δx = 0.5 cm for the steaks, and three nodes with a nodal spacing of Δr = 3.75 cm for the exposed portion of the de frosting plate. Also, use a time step of Δt = 4 s. Hint:First, determine the total amount of heat transfer needed to defrost the steaks, and then determine how long it will take to transfer that much heat.