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Calcualte the mass of stea that should be passed though 60 g of water at `20^(@)C`, such that the final temperature is `40^(@)C`. (take specific latent heat of steam is 2250 J `g^(-1)`). |
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Answer» let mass of stea be m gm heat given out by stea to form water at `100^(@)C=3xx2250" J "g^(-1)` heat given out by water at `100^(@)C` `=ms_(u)Delta theta=mxx4.2xx(100-40)` `=242" m "J" "g^(-1)` Total heat given out `=2250m+252m` `=2502" m J "g^(-1)` Heat gained by water at `20^(@)C=60xx4.2xx(40-20)` `=5040J` `therefore`Heat lost by a body=Heat gained by a body ltBrgt `2502m=5040` ltBrgt `m=(5040)/(2502)=2g` |
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