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10-1.Thermometry, Thermal Expansion and Calorimetry
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Steam at $100^o C$ is added slowly to $1400 \,\,gm$ of water at $16^o C$ until the temperature of water is raised to $80^o C$. The mass of steam required to do this is ($L_V =$ $540\,\,cal/gm$) ........... $gm$
A
$160$
B
$125 $
C
$250$
D
$320$
Solution
Let the mass of steam be $x$.
Heat lost by steam to come to $80^{\circ} \mathrm{C}$ from $100^{\circ} \mathrm{C}$ is
$Q_{\text {lost}}=x \times 540+x \times 1 \times(100-80)=560x$ $…(1)$
Heat gained by water $Q_{\text {gained }}=1400 \times 1 \times(80-16)=1400 \times 64$ $…(2)$
Equating $(1)$ and $( 2)$
$560 x=1400 \times 64$$…(2)$
$\therefore x=\frac{140 \times 64}{56}=160 \mathrm{gms}$
Standard 11
Physics
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