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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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A calorimeter (water equivalent = 25 g ) contains of 265 g of water at $\;30^{0}C\;$ . Steam is passed at $\;100^{0}C\;$ is passed which increases the temperature to $\;40^{0}C\;$ & the mass of the calorimeter and its contents by 2g . Given that the specific heat capacity of water is $\;1\;{calg^{-1}}^{0}C\;$ , the latent heat of vapourization of water is -

$(a)\;1390\;calg^{-1}\qquad(b)\;1265\;calg^{-1}\qquad(c)\;2900\;calg^{-1}\qquad(d)\;1450\;calg^{-1}$

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1 Answer

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Answer : (a) $\;1390\;calg^{-1}$
Explanation :
Mass of steam condensed = 2g
Heat lost in condensation = $\;Q_{L_{1}}=(2)L\;cal\;$
Heat lost when water cools from $\;100^{0}C\;$ to $\;40^{0}C\;$ =$\;2 \times 1 \times (100-40)=120 cal =Q_{L_{2}}$
Heat gained by water at $\;30^{0}C\;$ to rise upto $\;40^{0}C=Q_{g}=(265+25) \times (40-30)=2650\;cal\;$
Since $\;Q_{g}=Q_{L_{1}+Q_{L_{2}}} $
$=2900=2L+120$
$L=1390 \;calg^{-1}\;.$
answered Mar 12, 2014 by yamini.v
 

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