# 500g of water and 100g of ice at 0C are taken in a calorimeter who water equivalent is 40g. 10g of steam is at 100C is added to it. Then water in the calorimeter is: Latent heat of ice is 80cal/g and Latent heat of steam is 540cal/g

Dear Student,

Please find below the solution to the asked query:

Here, the energy released by the condensation of vapour is not sufficient to melt the whole ice. So, the amount of energy released by the water vapour, when it condensed and comes to *0 ^{0}C* is,

$E={m}_{vapour}{L}_{vapour}+{m}_{vapour}{S}_{vapour}\times 100\phantom{\rule{0ex}{0ex}}\Rightarrow E=10\times 540+10\times 1\times 100\phantom{\rule{0ex}{0ex}}\Rightarrow E=6400cal$

So, let

*m*mass of the ice is

*m'*. Therefore,

$m\text{'}{L}_{ice}=6400\Rightarrow m\text{'}=\frac{6400}{80}\Rightarrow m\text{'}=80gm$

Therefore, the total amount of water in the capillary tube is,

${m}_{0}=500+80+10\Rightarrow {m}_{0}=590gm$

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