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A thermally i...

Question

A thermally insulated container has 500 g of mercury in liquid state at its freezing point. The container is connected to a vacuum pump and air above the liquid (mercury) is pumped out the container. Due to this some of the liquid evaporates and the remaining liquid solidifies. Then find the mass of the solid that is found by solidification of liquid. Take latent heat of fusion , $L_{F} = 12500 J k g^{- 1}$ and Latent heat of vaporization $L_{B} = 290000 J K g^{- 1} .$

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Solution

Let the total mass of liquid = $M = \frac{1}{2} K g$
Let the mass of the liquid that is converted to solid = $m$
$L_{B} = 290000 J {k g}^{- 1} L_{F} = 12500 J {k g}^{- 1}$
$m (L_{F}) = (M - m) L_{B} \Rightarrow 12500 m = (0.5) (290000) - 290000 m \Rightarrow 302500 m = (0.5) (290000) m = 0.479 K g$

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