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10-1.Thermometry, Thermal Expansion and Calorimetry
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Steam at $100\,^oC$ is passed into $20\, g$ of water at $10\,^oC$. When water acquires a temperature of $80\,^oC$, the mass of water present will be ........ $g$
[Take specific heat of water $= 1\, cal\, g^{-1}\,^oC^{-1}$ and latent heat of steam $= 540\, cal\, g^{-1}$]
A
$24$
B
$31.5$
C
$42.5$
D
$22.5$
(AIPMT-2014)
Solution
Heat lost $=$ Heat gained
$\mathrm{mLv}+\mathrm{ms}_{\mathrm{w}} \Delta \theta=\mathrm{m}_{\mathrm{W}} \mathrm{s}_{\mathrm{w}} \Delta \theta$
$\Rightarrow \mathrm{m} \times 540+\mathrm{m} \times 1 \times(100-80)$
$=20 \times 1 \times(80-10)$
$\Rightarrow m=2.5 \mathrm{g}$
Total mass of water $=(20+2.5) g=22.5 g$
Standard 11
Physics
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