Thermal Properties Of Matter Ques 17
17. 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 :
(Given latent heat of fusion of ice $=3.36 \times$ $10^{5} $ $J $ $ kg^{-1}$.) :
[2012 M]
(a) $1.24 $ $J / m / s /{ }^{\circ} C$
(b) $1.29 $ $J / m / s /{ }^{\circ} C$
(c) $2.05 $ $J / m / s /{ }^{\circ} C$
(d) $1.02 $ $J / m / s /{ }^{\circ} C$
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Answer:
Correct Answer: 17.(a)
Solution:
- (a) According to condition,
Rate of heat given by steam $=$ Rate of heat taken by ice
where
$K=$ Thermal conductivity of the slab
$m=$ Mass of the ice
$L=$ Latent heat of melting/fusion
$A=$ Area of the slab
$\frac{d Q}{d t}=\frac{K A(100-0)}{1}=m \frac{d L}{d t}$
$\frac{K \times 100 \times 0.36}{0.1}=\frac{4.8 \times 3.36 \times 10^{5}}{60 \times 60}$
$K=1.24 $ $J / m / s /{ }^{\circ} C$
In electrical conduction
$I=\frac{d q}{d t}=\frac{V_1-V_2}{R}=\frac{\sigma A}{\ell}(V_1-V_2)$
In thermal conduction,
$H=\frac{d \theta}{d t}=\frac{\theta_1-\theta_2}{R}=\frac{k A}{\ell}(\theta_1-\theta_2)$
$K=$ Thermal conductivity of conductor.
BETA
