Question

A slab of stone of area 0.34 $m^2$ and thickness 10 cm is exposed on the lower face to steam at ${100}^{\circ }C$. A block of ice at $0^{\circ }C$ rests on the upper face of the slab. In one hour, 3.6 kg of ice is melted. Assume that the heat loss from the sides is negligible. The latent heat of fusion of ice is $3.4\times {10}^{5}Jk{g}^{-1}$. What is the thermal conductivity of the stone in units of $J{s}^{-1}{m}^{-1}{}^{\circ }{C}^{-1}$?

• A. 1.0
• B. 1.5
• C. 2.0
• D. 2.5

Answer 1

The correct option is A

1.0

Amount of heat flowing through the stone in 1 hour = mL, where m is the mass of ice melted and L its latent heat.
Now, time (t) = 1 hour = 3600 s and thickness d = 10 cm = 0.1 m.
The rate of flow of heat is $Q=\frac{mL}{t}$
$\therefore$ Thermal conductivity $k=\frac{Qd}{A({\theta }_2-{\theta }_1)}=\frac{mLd}{tA({\theta }_2-{\theta }_1)}$
$=\frac{3.6\times 3.4\times {10}^5\times 0.1}{3600\times 0.34\times (100-0)}=1J{s}^{-1}{m}^{-1}{}^{\circ }{C}^{-1}$
Hence, the correct choice is (a).