What is the enthalpy change (in J) for converting 9 g of $\mathrm{H}_2 \mathrm{O}$ (l) at $+10^{\circ}…

What is the enthalpy change (in J) for converting 9 g of $\mathrm{H}_2 \mathrm{O}$ (l) at $+10^{\circ} \mathrm{C}$ to $\mathrm{H}_2 \mathrm{O}$ (l) at $+20^{\circ} \mathrm{C}$ ? $\left(\mathrm{C}_{\mathrm{p}}\left(\mathrm{H}_2 \mathrm{O}\right.\right.$ (1)) $\left.=75 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\left(\right.$ density of $\left.\mathrm{H}_2 \mathrm{O}(\mathrm{l})=1 \mathrm{~g} \mathrm{~mL}^{-1}\right)$
  1. 750
  2. 75
  3. 37.5
  4. 375

Solution

We know that, $\Delta H=C_p \Delta T$
Given $\begin{aligned} & \mathrm{C}_{\mathrm{p}}=75 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{k}^{-1} \\ & \mathrm{~T}_1=283 \mathrm{k} \\ & \mathrm{~T}_2=293 \mathrm{k} \end{aligned}$ $\begin{array}{r} \operatorname{mole}(\mathrm{n})=\frac{\text { Weight of } \mathrm{H}_2 \mathrm{O}}{\text { Molecular weight of } \mathrm{H}_2 \mathrm{O}} \\ \Rightarrow \frac{9}{18}=\frac{1}{2} \mathrm{~mol} \end{array}$ $\begin{aligned} & \therefore \quad \Delta \mathrm{H}=75 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{k}^{-1}(293-283) \mathrm{k} \times \frac{1}{2} \mathrm{~mol} \\ & \Delta \mathrm{H}=\frac{750}{2} \mathrm{~J} \\ & \Delta \mathrm{H}=375 \mathrm{~J} \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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