A fully charged capacitor has a capacitance '$$C$$'. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity '$$s$$' and mass '$$m$$'. If the temperature of the block is raised by '$$\Delta T$$'. The potential difference $$V$$ across the capacitance is

By the principle of conservation of energy, since the block is thermally insulated, the total electrical energy stored in the fully charged capacitor is completely dissipated as heat across the resistance wire, which is then absorbed by the block.

The initial electrical energy stored in the capacitor is:

$$ E = \frac{1}{2} C V^2 $$

The heat energy gained by the block to raise its temperature by $$\Delta T$$ is given by the calorimetry formula:

$$ Q = m s \Delta T $$

Equating the electrical energy dissipated to the heat energy absorbed ($$E = Q$$):

$$ \frac{1}{2} C V^2 = m s \Delta T $$

Now, solve for the potential difference $$V$$:

$$ V^2 = \frac{2 m s \Delta T}{C} $$

$$ V = \sqrt{\frac{2 m s \Delta T}{C}} $$