Q12.19P - Chemistry: Molecular Nature Of Matter And Change

Question

From the data below, calculate the total heat (in J) needed to convert 22.00 g of ice at -6.00${}^o\mathbf{C}$ to liquid water at 0.500 ${}^o\mathbf{C}$: M.P at 1 atm: 0.0 ${}^o\mathbf{C}$; ${\mathbf{\Delta H}}_{\mathrm{fus}}$ : 6.02 kJ/mol; ${\mathbf{Cl}}_{\mathrm{iquid}}$ : 4.21 J/g${}^o\mathbf{C}$; ${\mathbf{C}}_{\mathrm{solid}}$: 2.09 J/g${}^o\mathbf{C}$ 

Step-by-Step Solution

Answer

The heat required is 7679.96 J in the process. 

Step-1: Heat needed to convert 22.00 g of ice at -6.00 o C to 0 o C

The heat required to convert 22.00 g of ice at -6.00 ${}^{o}C$to 0 ${}^{o}C$is calculated as:

 $\Delta H_1=C_{solid}(J/g^{o}C)*Weight\left(g\right)*\Delta T{(}^{o}C)$

$\Delta H_1=2.09*22*6=275.88J$

Step-2: Heat needed to convert 22.00 g of ice to water

The heat required to convert 22.00 g of ice to water is 

$\begin{array}{c}\Delta H_2=6020\frac{J}{mol}\\ =6020*\frac{22}{18}J\\ \Delta H_2=7357.77J\end{array}$

Step-3: Heat needed to convert 22.00 g of water at 0.00 o C to 0.50 o C

The heat required to convert 22.00 g of water at 0.00${}^{o}C$ to 0.50${}^{o}C$ is calculated as:

$\Delta H_3=C_{solid}(J/g^{o}C)*Weight\left(g\right)*\Delta T{(}^{o}C)$

$\Delta H_3=4.21*22*0.5=46.71J$
 

Hence, the total heat energy required in the process is 

 $\begin{array}{c}\Delta H=\Delta H_1+\Delta H_2+\Delta H_3\\ =275.88+7357.77+46.71\\ \Delta H=7679.96J\end{array}$