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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