100 ,gm ગ્રામ પાણીનું તાપમાન 24^{circ} C થી વ | vedclass

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Question

(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 fi

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(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 $=536 	imes m$ calne

Final Temp of solution $=m 	imes$ specific heat of water $	imes$ fall of temp

$= m 	imes 1 	imes(100-90)$

$= m 	imes 1 	imes 10$

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

$24^{\circ}\,C -90^{\circ}\,C$

$=(100+m) 	imes 1 	imes(90-24)$

$=(100+ m ) 	imes 66$ calories

heat gained $=$ Heat lost

$(100+ m ) 	imes 66=546\,m$

$6600+66\,m =546\,m$

$600=486\,m$

$m =13.75\,g$ of steam $\simeq 13\,gm$

$100 \,gm$ ગ્રામ પાણીનું તાપમાન $24^{\circ} C$ થી વધારીને $90^{\circ} C$ કરવું હોય તો તેમાંથી .......... ગ્રામ વરાળ પસાર કરવી પડે ?

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