ChemistryQ151–151 of 177 questions
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ChemistryEasyMCQTS EAMCET · 2012
स्थिर दाब पर जल की मोलर ऊष्मा धारिता $(C_p)$ $75 \ J \ K^{-1} \ mol^{-1}$ है। जब $100 \ g$ जल को $1 \ kJ$ ऊष्मा दी जाती है,तो उसके तापमान में होने वाली वृद्धि ($K$ में) क्या होगी?
A
$2.4$
B
$0.24$
C
$1.3$
D
$0.13$
Solution
(A) स्थिर दाब पर दी गई ऊष्मा का सूत्र $\Delta H = n C_p \Delta T$ है।
दिया गया है:
दी गई ऊष्मा $(q)$ = $1 \ kJ = 1000 \ J$.
जल का द्रव्यमान $(m)$ = $100 \ g$.
जल का मोलर द्रव्यमान $(M_{H_2O})$ = $18 \ g \ mol^{-1}$.
मोलों की संख्या $(n)$ = $\frac{m}{M} = \frac{100}{18} \ mol$.
$C_p = 75 \ J \ K^{-1} \ mol^{-1}$.
मान रखने पर:
$1000 = (\frac{100}{18}) \times 75 \times \Delta T$.
$\Delta T = \frac{1000 \times 18}{100 \times 75} = \frac{180}{75} = 2.4 \ K$.
दिया गया है:
दी गई ऊष्मा $(q)$ = $1 \ kJ = 1000 \ J$.
जल का द्रव्यमान $(m)$ = $100 \ g$.
जल का मोलर द्रव्यमान $(M_{H_2O})$ = $18 \ g \ mol^{-1}$.
मोलों की संख्या $(n)$ = $\frac{m}{M} = \frac{100}{18} \ mol$.
$C_p = 75 \ J \ K^{-1} \ mol^{-1}$.
मान रखने पर:
$1000 = (\frac{100}{18}) \times 75 \times \Delta T$.
$\Delta T = \frac{1000 \times 18}{100 \times 75} = \frac{180}{75} = 2.4 \ K$.
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