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$100 \,gm$ ગ્રામ પાણીનું તાપમાન $24^{\circ} C$ થી વધારીને $90^{\circ} C$ કરવું હોય તો તેમાંથી .......... ગ્રામ વરાળ પસાર કરવી પડે ?
$20$
$15$
$13$
$18$
Solution
(c)
Let mass of steam required $= m gm$
each gram of steam of condensing release 536 calories of heat steam condense at $100^{\circ} C$ cools finally to $90^{\circ} C$
Heat released by $m$ gm of steam on condensing $=536 \times m$ calne
Final Temp of solution $=m \times$ specific heat of water $\times$ fall of temp
$= m \times 1 \times(100-90)$
$= m \times 1 \times 10$
heat released $=536\, m +10\, m$
$546 m$ calories of heat
Heat required to raised the temp of $100 gm$ of water at $24^{\circ} C + m g m$ of condensed steam from
$24^{\circ}\,C -90^{\circ}\,C$
$=(100+m) \times 1 \times(90-24)$
$=(100+ m ) \times 66$ calories
heat gained $=$ Heat lost
$(100+ m ) \times 66=546\,m$
$6600+66\,m =546\,m$
$600=486\,m$
$m =13.75\,g$ of steam $\simeq 13\,gm$
