ChemistryQ151–151 of 177 questions
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ChemistryEasyMCQTS EAMCET · 2012
Molar heat capacity $(C_p)$ of water at constant pressure is $75 \ J \ K^{-1} \ mol^{-1}$. The increase in temperature (in $K$) of $100 \ g$ of water when $1 \ kJ$ of heat is supplied to it is
A
$2.4$
B
$0.24$
C
$1.3$
D
$0.13$
Solution
(A) The formula for heat supplied at constant pressure is $\Delta H = n C_p \Delta T$.
Given:
Heat supplied $(q)$ = $1 \ kJ = 1000 \ J$.
Mass of water $(m)$ = $100 \ g$.
Molar mass of water $(M_{H_2O})$ = $18 \ g \ mol^{-1}$.
Number of moles $(n)$ = $\frac{m}{M} = \frac{100}{18} \ mol$.
$C_p = 75 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values:
$1000 = (\frac{100}{18}) \times 75 \times \Delta T$.
$\Delta T = \frac{1000 \times 18}{100 \times 75} = \frac{180}{75} = 2.4 \ K$.
Given:
Heat supplied $(q)$ = $1 \ kJ = 1000 \ J$.
Mass of water $(m)$ = $100 \ g$.
Molar mass of water $(M_{H_2O})$ = $18 \ g \ mol^{-1}$.
Number of moles $(n)$ = $\frac{m}{M} = \frac{100}{18} \ mol$.
$C_p = 75 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values:
$1000 = (\frac{100}{18}) \times 75 \times \Delta T$.
$\Delta T = \frac{1000 \times 18}{100 \times 75} = \frac{180}{75} = 2.4 \ K$.
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