Chapter 17, Problem 59GP

To determine

The potential difference across the plates.

Answer to Problem 59GP

  $620V$

Explanation of Solution

Given data:

Mass, $m=2.5kg$

Capacitance, $C=4.0F$

Temperature, $T=95°C-21°C$

Formula used:

It is known that potential energy stored in a capacitor:

  $PE=Energy=\frac{1}{2}QV$ .

Since, $Q=CV$ , then

  $PE=\frac{1}{2}QV=\frac{1}{2}C{V}^2$

Where,

  $C$ is the capacitance.

  $\text{V}$ is the potential difference.

  $Q$ is the charge on each plate.

  $\text{Q}\text{=}\text{mcΔT}$

Where,

  $\text{Q}$ is the amount of heat flowing.

  $\text{m}$ is mass.

  $\text{c}$ is specific heat capacity.

  $\Delta \text{T}$ is the final temperature.

Calculation:

The energy in the capacitor is the heat energy absorbed by the water. So,

  $\begin{array}{l}PE=Q_{Heat}\\ \Rightarrow \frac{1}{2}C{V}^2=mc\Delta T\\ \Rightarrow V=\sqrt{\frac{2mc\Delta T}{C}}\mathrm{...}\left(1\right)\end{array}$

Substitute the given data into equation $\left(1\right)$ to calculate the potential difference:

  $V=\sqrt{\frac{2\left(2.5kg\right)\left(4186\frac{J}{kg\cdot °C}\right)\left(95°C-21°C\right)}{4.0F}}$

  $=622V\approx 620V$

Conclusion:

The required value is $620V$ .