If 1μF capacitor connected to 1kV battery, then it's charge will be 1mC?

Question

Work done by the battery
Work = UQ = 1 Joule
Energy strored in capacitor according to my textbook
Energy = 1/2 CU² = 1/2 Joule

Where did 1/2 Joule of energy disappear?

Answer 1

In heating the battery through its internal resistance. At the instant of connection, the battery voltage is forced to zero and the internal resistance limits the initial current.