Question

A slab of stone of area 0.36 m2 and thickness 0.1 m is exposed on the lower surface to steam at 100${}^{\circ }$C. A block of ice at 0${}^{\circ }$C rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of slab is
(Given latent heat of fusion of ice = $3.36\times 10^{5}Jk{g}^{-1})$

• A. $1.24J/m/s/\circ C$
• B. $1.29J/m/s/\circ C$
• C. $2.05J/m/s/\circ C$
• D. $1.02J/m/s/\circ C$

Answer 1

Answer: (A)

Heat flows through the slab in t s is
$Q=\frac{KA(T_1-T_2)t}{L}=\frac{K\times 0.36\times (100-0)\times 3600}{0.1}$
$\frac{K\times 0.36\times 100\times 3600}{0.1}$ .....................(i)
So ice melted by this heat is
$m_{ice}=\frac{Q}{L^f}$ . ...............(ii)
or $Q=m_{ice}{L}_f=4.8\times \times {10}^5$
From (i) and (ii), we get
$\frac{K\times 0.36\times (100-0)\times 3600}{0.1}=4.8\times 3.36\times {10}^5$
$K=\frac{4.8\times 3.36\times {10}^5\times 0.1}{0.36\times 100\times 3600}=1.24J/m/s/\circ C$