The heat of solution of anhydrous $CuS{O}_4$ is -15.9k.cal. and that of $CuS{O}_4.5H_{2}O$ is 2.8k cal. Then the heat of hydration of $CuS{O}_4$. In Kcal is

${\text{CuSO}}_{\text{4}}\left(\text{s}\right)\to {{\text{\hspace{0.17em}CuSO}}_{\text{4}}}_{\left(\text{eq}\right)}\text{\_\_\_\_\_}\left(\text{1}\right)$

${\text{ΔH}}_{\text{1}}\text{\hspace{0.17em}=\hspace{0.17em}-15.9}$

${\text{Cu\hspace{0.17em}SO}}_{\text{4}}{\text{.\hspace{0.17em}5\hspace{0.17em}\hspace{0.17em}H}}_{\text{2}}\text{O\hspace{0.17em}}\to {\text{\hspace{0.17em}CuSO}}_{\text{4\hspace{0.17em}\hspace{0.17em}eq}}+{\text{\hspace{0.17em}5\hspace{0.17em}H}}_{\text{2}}\text{O}$

${\text{ΔH}}_{\text{2}}\text{=\hspace{0.17em}2.8\hspace{0.17em}\hspace{0.17em}\_\_\_\_\_\_}\left(\text{2}\right)$

revers  equation (2)

add (2) & (2)

${\text{Cu\hspace{0.17em}SO}}_{\text{4\hspace{0.17em}}\left(\text{s}\right)}\to {\text{\hspace{0.17em}\hspace{0.17em}CuSO}}_{\text{4\hspace{0.17em}}\left(\text{aq}\right)}$

$\underline{{\text{Cu\hspace{0.17em}SO}}_{\text{4\hspace{0.17em}}\left(\text{aq}\right)}+{\text{\hspace{0.17em}5H}}_{\text{2}}\text{O\hspace{0.17em}\hspace{0.17em}}\to {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}CuSO}}_{\text{4}}.{\text{5H}}_{\text{2}}\text{O}}$

${\text{CuSO}}_{\text{4\hspace{0.17em}\hspace{0.17em}}\left(\text{s\hspace{0.17em}}\right)}{\text{\hspace{0.17em}+\hspace{0.17em}5H}}_{\text{2}}O\to {\text{\hspace{0.17em}\hspace{0.17em}Cu\hspace{0.17em}SO}}_{\text{4}}{\text{\hspace{0.17em}\hspace{0.17em}.\hspace{0.17em}\hspace{0.17em}5H}}_{\text{2}}\text{O}$

- 15 . 9 + (-2.8)

 - 18.7 K.cal