The amount of heat trsnsferred though the slab to the ice in one hour is
`Q=(4.8xx10^(-3)kg)xx(3.36xx10^(5)J kg^(-1)` . `=4.8xx336J` . Using the equation `Q=(KA(theta_(1)-theta_(2))t)/(x)` , `4.8xx336J=(k(3600cm^(2))(100^(@)C)xx(3600s))/(10cm)` . Or, `K=1.24xx10^(-3)Wm^(-1)`^(@)C^(-1)` .