Question

A huge 4.0-F capacitor has enough stored energy to heat 2.8 kg of water from 21°C to 95°C. What is the potential difference across the plates?

Short answer

The potential difference across the plates of 4.0 F capacitor is 658.5 V.

Step-by-step solution

Understanding of potential energy stored inside a capacitor

The capacitor is a charge and energy storage device. The electric potential energy stored in a capacitor relies on the capacitance and potential difference across the plates.

The electric potential energy stored in a capacitor is given by the following:

\(PE = \frac{1}{2}C{V2}\) … (i)

Here, C is the capacitance, and V is the potential difference.

Given information

The capacitance of the capacitor is \(C = 4.0\;{\rm{F}}\).

The mass of water is m = 2.8 kg.

The initial temperature of the water is \(T = 21\circ \;{\rm{C}}\).

The final temperature of the water is \(T' = 95\circ {\rm{C}}\).

Specific heat of water is \(c = 4186\;{\rm{J/kg}} \cdot {\rm{\circ C}}\).

Determination of potential difference across the plates of the capacitor

The amount of heat energy required by the water to increase its temperature from T to \(T'\) is as follows:

\(Q = mc\left( {T' - T} \right)\) … (ii)

The energy stored inside the capacitor is given to the water to increase its temperature. From equations (i) and (ii),

\(\begin{array}{c}PE = Q\\\frac{1}{2}C{V2} = mc\left( {T' - T} \right)\\V = \sqrt {\frac{{2mc\left( {T' - T} \right)}}{C}} \end{array}\)

Substitute the values in the above expression.

\(\begin{array}{c}V = \sqrt {\frac{{2\left( {2.8\;{\rm{kg}}} \right)\left( {4186\;{\rm{J/kg}} \cdot {\rm{\circ C}}} \right)\left( {95\circ {\rm{C}} - 21\circ {\rm{C}}} \right)}}{{4.0\;{\rm{F}}}}} \\ = \sqrt {433669.6} \;{\rm{V}}\\ = 658.5\;{\rm{V}}\end{array}\)

Thus, the potential difference across the plates of the capacitor is 658.5 V.