ChemistryQ151–151 of 177 questions
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151
ChemistryEasyMCQTS EAMCET · 2012
અચળ દબાણે પાણીની મોલર ઉષ્મા ધારિતા $(C_p)$ $75 \ J \ K^{-1} \ mol^{-1}$ છે. જ્યારે $100 \ g$ પાણીને $1 \ kJ$ ઉષ્મા આપવામાં આવે ત્યારે તેના તાપમાનમાં થતો વધારો ($K$ માં) કેટલો હશે?
A
$2.4$
B
$0.24$
C
$1.3$
D
$0.13$
Solution
(A) અચળ દબાણે આપેલી ઉષ્માનું સૂત્ર $\Delta H = n C_p \Delta T$ છે.
આપેલ છે:
આપેલી ઉષ્મા $(q)$ = $1 \ kJ = 1000 \ J$.
પાણીનું દળ $(m)$ = $100 \ g$.
પાણીનું આણ્વીય દળ $(M_{H_2O})$ = $18 \ g \ mol^{-1}$.
મોલની સંખ્યા $(n)$ = $\frac{m}{M} = \frac{100}{18} \ mol$.
$C_p = 75 \ J \ K^{-1} \ mol^{-1}$.
કિંમતો મૂકતા:
$1000 = (\frac{100}{18}) \times 75 \times \Delta T$.
$\Delta T = \frac{1000 \times 18}{100 \times 75} = \frac{180}{75} = 2.4 \ K$.
આપેલ છે:
આપેલી ઉષ્મા $(q)$ = $1 \ kJ = 1000 \ J$.
પાણીનું દળ $(m)$ = $100 \ g$.
પાણીનું આણ્વીય દળ $(M_{H_2O})$ = $18 \ g \ mol^{-1}$.
મોલની સંખ્યા $(n)$ = $\frac{m}{M} = \frac{100}{18} \ mol$.
$C_p = 75 \ J \ K^{-1} \ mol^{-1}$.
કિંમતો મૂકતા:
$1000 = (\frac{100}{18}) \times 75 \times \Delta T$.
$\Delta T = \frac{1000 \times 18}{100 \times 75} = \frac{180}{75} = 2.4 \ K$.
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