10-2.Transmission of Heat
medium

A slab of stone of area $0.36\;m ^2$ 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 .......... $J / m / s /{ }^{\circ} C$ (Given latent heat of fusion of ice $=3.36 \times 10^5\; J kg ^{-1}$)

A

$1.02$

B

$1.29$

C

$1.24$

D

$2.05$

(AIPMT-2012)

Solution

$Q=m L_f$

$\frac{K A}{L}\left(T_1-T_2\right) t=m L_f$

$K=\frac{m L_f L}{A\left(T_1-T_2\right) t}$

$K=\frac{4.8 \times 3.36 \times 10^5 \times 0.1}{0.36 \times 100 \times 3600}$

$K=\frac{4.8 \times 3.36}{0.36 \times 36}=1.24\; J / m / s ^{\circ} C$

Standard 11
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.