What is the enthalpy change (in J) for conver | vedclass

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(D) The formula for enthalpy change during heating is $\Delta H = n C_p \Delta T$. Given: $C_p = 75 \ J \ mol^{-1} \ K^{-1}$ $T_1 = 10^{\circ}C = 283

Vedclass · Accepted Answer

$375$

Vedclass · Answer

(D) The formula for enthalpy change during heating is $\Delta H = n C_p \Delta T$. Given: $C_p = 75 \ J \ mol^{-1} \ K^{-1}$ $T_1 = 10^{\circ}C = 283 \ K$ $T_2 = 20^{\circ}C = 293 \ K$ $\Delta T = 293 - 283 = 10 \ K$ Number of moles $(n)$ = $\frac{	ext{mass}}{	ext{molar mass}} = \frac{9 \ g}{18 \ g \ mol^{-1}} = 0.5 \ mol$. Substituting the values: $\Delta H = 0.5 \ mol 	imes 75 \ J \ mol^{-1} \ K^{-1} 	imes 10 \ K$ $\Delta H = 375 \ J$.

What is the enthalpy change (in $J$) for converting $9 \ g$ of $H_2O(l)$ at $+10^{\circ}C$ to $H_2O(l)$ at $+20^{\circ}C$? $(C_p(H_2O(l)) = 75 \ J \ mol^{-1} \ K^{-1})$

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