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A tube leads from a flask in which water is boiling under atmospheric pressure to a calorimeter. The mass of the calorimeter is 150 g, its specific heat capacity is `0.1 cal//g//^@C`, and it contains originally 340g of water at `15^@C`. Steam is allowed to condense in the colorimeter until its temperature increases to `71^@C`, after which total mass of calorimeter and contents are found to be 525g. Compute the heat of condensation of steam. |
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Answer» Mass of calorimeter and contents before passing steam `=(150+340)=490g` mass after passing steam`=525g` `implies`mass of steam which condenses`=(525-490)g=35g` Let L`=` latent heat of steam. Heat lost by steam`=`heat gained by water`+`heat gained by calorimeter `35L+35xx1(100-71)=340xx1xx(71-15)+150` `xx0.1xx(71-15)` `implies=539 cal//g` |
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