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 = $3.36 \times 10^5 J kg^{-1})$

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

Answer 1

Answer: (A)

$1.24J/m/s / \circ C$

Answer 2

Heat flows through the slab in t s is $Q= \frac{KA(T_1-T_2)t}{L} = \frac{K\times 0.36 \times (100-0)\times 3600}{ 0.1}$ $\frac{ K\times 0.36 \times 100 \times 3600 }{ 0.1}$ \hspace20mm .....................(i) So ice melted by this heat is $m_{ice} = \frac{Q}{L^f}$ \hspace20mm . ...............(ii) or $Q = m_{ice} L_f = 4.8 \times \times {10}^5$ From (i) and (ii), we get $\frac{ K \times 0.36 \times (100-0) \times 3600}{0.1} = 4.8\times3.36\times {10}^5$ $K= \frac{ 4.8 \times 3.36 \times{10}^5 \times 0.1}{ 0.36\times100\times3600} =1.24 J/m/s / \circ C$