Given data

• Heat of solution of anhydrous copper sulfate: $\Delta H_{\text{sol}}(\mathrm{CuSO_4}) = -70\ \text{kJ mol}^{-1}$

• Heat of solution of hydrated copper sulfate: $\Delta H_{\text{sol}}(\mathrm{CuSO_4\cdot 5H_2O}) = +12\ \text{kJ mol}^{-1}$

• Heat of hydration: $\mathrm{CuSO_4 + 5H_2O \to CuSO_4\cdot 5H_2O}, \quad \Delta H = -x\ \text{kJ}$

Write the dissolution processes

For anhydrous salt: $\mathrm{CuSO_4(s) \to CuSO_4(aq)} \qquad \Delta H = -70\ \text{kJ mol}^{-1}$

For hydrated salt: $\mathrm{CuSO_4\cdot 5H_2O(s) \to CuSO_4(aq) + 5H_2O(l)} \qquad \Delta H = +12\ \text{kJ mol}^{-1}$

Apply Hess's law

Hydration reaction is: $\mathrm{CuSO_4(s) + 5H_2O(l) \to CuSO_4\cdot 5H_2O(s)} \qquad \Delta H = -x$

Reverse of this reaction is: $\mathrm{CuSO_4\cdot 5H_2O(s) \to CuSO_4(s) + 5H_2O(l)} \qquad \Delta H = +x$

Now add this reversed hydration reaction to dissolution of anhydrous salt:

$\mathrm{CuSO_4\cdot 5H_2O(s) \to CuSO_4(s) + 5H_2O(l)} \qquad \Delta H = +x$ $\mathrm{CuSO_4(s) \to CuSO_4(aq)} \qquad \Delta H = -70$

Adding: $\mathrm{CuSO_4\cdot 5H_2O(s) \to CuSO_4(aq) + 5H_2O(l)}$ $\Delta H = x - 70$

But this is exactly the heat of solution of $\mathrm{CuSO_4\cdot 5H_2O}$, which is given as $+12\ \text{kJ mol}^{-1}$.

Therefore, $x - 70 = 12$

Solve for $x$

$x = 82$

Final answer

$\boxed{82}$