A slab of stone of area 0.36m² and thickness 0.1m is exposed on the lower surface to steam at 100°C. A block of ice at 0°C rests on the upper surface of the slab. In one hour 4.8kg of ice is melted. The thermal conductivity of the slab is : (Given latent heat of fusion of ice = 3.36 × 10⁵ J/kg)

From Fourier's law we have,
dQ/dt = KA(T₁ - T₂)/L
Q = KA(T₁ - T₂)t/L
Since Q is the heat used to melt the ice, Q = mLf
KA(T₁ - T₂)t/L = mLf
K = mLf(L)/A(T₁ - T₂)t
Given:
m = 4.8 kg
Lf = 3.36 × 10⁵ J/kg
A = 0.36 m²
T₁ = 100°C
T₂ = 0°C
L = 0.1 m
t = 1 hour = 3600 s
Substituting the values,
K = (4.8 kg)(3.36 × 10⁵ J/kg)(0.1 m)/(0.36 m²)(100°C)(3600 s)
K = 1.24 J/m/s/°C