1.

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)`).

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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