A thermos-flask of negligible heat capacity contains 100 g of ice. Calculate the amount of water in the flask after condensation.
(Given that: Latent heat of ice $= 336 J g^{- 1}$ , latent heat of steam $= 2260 J g^{- 1}$ , specific heat capacity of water $= 4.2 J g^{- 1} C^{- 1}$ )

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Solution

Let mass of steam passed in the flask be m grams. Then, Heat energy released by the steam, Q = mL
$= m g r a m \times 2260 J / g = 2260 m J$
Since, Heat given = Heat taken, therefore,
$2260 m = 22600 \Rightarrow m = \frac{33600}{2260} = 14.87 g$
$∴$ Total amount of water in the flask after condensation $= 100 g + 30 g + 14.87 g = 144.87 g$

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