Gujarati
Hindi
10-1.Thermometry, Thermal Expansion and Calorimetry
hard

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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