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

Thermodynamics

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

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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}}} $

$=2900=2L+120$

$L=1390 \;calg^{-1}\;.$

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