The heat of hydration of CuSO4 to CuSO4 · 5H2O is given by the difference between the heat of solution of anhydrous CuSO4 and CuSO4 · 5H2O.
Heat of hydration = Heat of solution of anhydrous CuSO4 − Heat of solution of CuSO4 · 5H2O.
Substitute the values:
\( x = \left| -70 \, \text{kJ mol}^{-1} - \left( +12 \, \text{kJ mol}^{-1} \right) \right| \).
\( x = \left| -70 - 12 \right| = \left| -82 \right| = 82 \).
Final Answer: 82 kJ.
Match List - I with List - II.

Consider the following statements:
(A) Availability is generally conserved.
(B) Availability can neither be negative nor positive.
(C) Availability is the maximum theoretical work obtainable.
(D) Availability can be destroyed in irreversibility's.
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to:
A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is: