The charge gets shared when two conductors of different potentials are brought close to each other. During this process, some amount of energy is lost as heat energy.

In order to calculate the total loss of energy during this process, consider two capacitors having capacitance C₁ and C₂ having distinct potentials V₁ and V₂ respectively. Now we know that, the charge always flows from a capacitor of high potential to a capacitor of low potential.
Hence, the common potential is given as:
V = (C₁V₁ + C₂ V₂)  / C₁ + C₂  .......... (1)
Total energy of the system before contact will be:
E₁ = 1/2C₁V₁² + 1/2 C₂ V₂²   ........... (2)

Total energy of the system after contact will be:
E₂ = 1/2 (C₁ + C₂) V²
E₂ = [ 1/2 (C₁ + C₂) ] . [ (C₁V₁ + C₂ V₂)  / (C₁ + C₂) ]²  (Using equation (1))
E₂ = [ 1/2 (C₁ + C₂) ] . [ (C₁V₁ + C₂ V₂)² / (C₁ + C₂)² ]
E₂ = (C₁V₁ + C₂ V₂)² / 2 (C₁ + C₂)   .......... (3)
Subtracting equation (2) and equation (3), we get:
E₁ - E₂ = [1/2 C₁V₁² + 1/2 C₂ V₂² ] - [ (C₁V₁ + C₂ V₂)² / 2 (C₁ + C₂) ]
E₁ - E₂ = [ C₁V₁² (C₁ + C₂) + C₂ V₂² (C₁ + C₂) - (C₁V₁ + C₂ V₂)² ] / 2 (C₁ + C₂)
E₁ - E₂ =C₁²V₁²+C₁C₂V₁²+C₁C₂V₂²+C₂²V₂²-(C₁²V₁²+C₂²V₂² +2C₁C₂V₁V₂)/2(C₁+C₂)
E₁ - E₂ =C₁²V₁²+C₁C₂V₁²+C₁C₂V₂²+C₂²V₂²- C₁²V₁²-C₂²V₂² - 2C₁C₂V₁V₂ /2(C₁ + C₂)
E₁ - E₂ = C₁C₂ (V₁ - V₂)² / 2(C₁ + C₂)
The above equation shows that E₁ - E₂ is greater then zero.
Or E₁ is greater then E₂
Or E₂ is less then E₁

Hence, it clearly shows that some amount of energy is lost as heat energy.