Question

A slab of stone of area 0.36m² and thickness 0.1m is exposed on the lower surface to steam at 373 K. A block of ice at 273 k rests on the upper surface of the slab. If in one hour, 4.8kg of Ice is melted, calculate the conductivity of the stone. Latent heat of Ice = 3.35 x 10⁵ JK^-1.

Answer 1

A = 0.36m²,

d = 0.1m

θ₁ = 373 K

θ₂ = 273 K

t = 1 hr = 3600s,

m = 4.8kg,

K = ?

L = 3.35 × 10⁵ Jkg^-1

Heat conducted through the rod = Heal used in melting ice

= 1.241 Wm^-1 K^-1.