Thermal Properties of Matter - Result Question 23
23. 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 $=3.36 \times$ $10^{5} J kg^{-1}$.) :
[2012M]
(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: 23. (a)
Solution:
- (a) According to condition,
$ 100^{\circ} C \text{ (steam) } $
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. 25 .
