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Amount-of-heat-required-to-change-1-g-ice-at-0-C-to-1-g-steam-at-100-C-is-1-616-cal-2-12-kcal-3-717-cal-4-none-of-these-




Question Number 23981 by Tinkutara last updated on 10/Nov/17
Amount of heat required to change 1 g  ice at 0°C to 1 g steam at 100°C is  (1) 616 cal  (2) 12 kcal  (3) 717 cal  (4) none of these.
$$\mathrm{Amount}\:\mathrm{of}\:\mathrm{heat}\:\mathrm{required}\:\mathrm{to}\:\mathrm{change}\:\mathrm{1}\:{g} \\ $$$$\mathrm{ice}\:\mathrm{at}\:\mathrm{0}°\mathrm{C}\:\mathrm{to}\:\mathrm{1}\:{g}\:\mathrm{steam}\:\mathrm{at}\:\mathrm{100}°\mathrm{C}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{616}\:\mathrm{cal} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{12}\:\mathrm{kcal} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{717}\:\mathrm{cal} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{none}\:\mathrm{of}\:\mathrm{these}. \\ $$
Answered by ajfour last updated on 10/Nov/17
Q=mL_f +mc△T+mL_(vap)      ≈(80+100+540)cal =720 cal .
$${Q}={mL}_{{f}} +{mc}\bigtriangleup{T}+{mL}_{{vap}} \\ $$$$\:\:\:\approx\left(\mathrm{80}+\mathrm{100}+\mathrm{540}\right){cal}\:=\mathrm{720}\:{cal}\:. \\ $$
Commented by Tinkutara last updated on 10/Nov/17
Should I learn those values? Or that  should be given in question?
$$\mathrm{Should}\:\mathrm{I}\:\mathrm{learn}\:\mathrm{those}\:\mathrm{values}?\:\mathrm{Or}\:\mathrm{that} \\ $$$$\mathrm{should}\:\mathrm{be}\:\mathrm{given}\:\mathrm{in}\:\mathrm{question}? \\ $$
Commented by ajfour last updated on 10/Nov/17
No harm remembering, but  i think, they are given in Q. paper.
$${No}\:{harm}\:{remembering},\:{but} \\ $$$${i}\:{think},\:{they}\:{are}\:{given}\:{in}\:{Q}.\:{paper}. \\ $$
Commented by Tinkutara last updated on 10/Nov/17
OK Thanks.
$$\mathrm{OK}\:\mathrm{Thanks}. \\ $$

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