Question

Problem #11 (10 mins) A concrete slab is 13.5 cm thick and has an area of 5.00 m2. Electric heating coils are installed under the slab to melt the ice on the surface in the winter months. What minimum power must be supplied to the coils to maintain a temperature difference of 22.5°C between the bottom of the slab and its surface? Assume all the energy transferred is through the slab, and the thermal conductivity of the concrete is 1.3 J/s m C.

Answer 1

Answer (11) : Power = 1083.33 Watt

The equation you want to use is:
q = k A dT / s

where

q = heat transferred per unit time (W, Btu/hr)
A = heat transfer area (m2)
k = thermal conductivity of the material (W/m.K)
dT = temperature difference across the material (oC)
s = material thickness (m)

q = (1.3 x 5 x 22.5) / .135

q = 1083.33 J/s or 1083.33 W