Chapter 17, Problem 56GP

(a)

To determine

The potential difference.

Answer to Problem 56GP

  $1.1\times {10}^6\text{V}$

Explanation of Solution

Given data:

Charges, $\text{q}\text{=}\text{4}\text{.0}\text{C}$

Energy is $\text{4}\text{.2}{\text{×10}}^{\text{6}}\text{J}$

Formula used:

  $\Delta \text{V=}\frac{\Delta \text{PE}}{\text{q}}$

Where,

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

  $\Delta \text{PE}$ is change in potential energy.

  $\text{q}$ is charge.

Calculation:

The energy is related to the charge and the potential difference is calculated by $\Delta \text{V=}\frac{\Delta \text{PE}}{\text{q}}\mathrm{...}\left(1\right)$ .

Substitute the given values in equation $\left(1\right)$

  $\begin{array}{c}\text{ΔV=}\frac{\text{4}\text{.2}{\text{×10}}^{\text{6}}\text{J}}{\text{4}\text{.0}\text{C}}\\ \text{=1}\text{.05}{\text{×10}}^{\text{6}}\text{V}\\ =1.1\times {10}^6\text{V}\end{array}$

(b)

To determine

The quantity of boiled and vaporized water.

Answer to Problem 56GP

  $\text{1}\text{.6}\text{kg}$

Explanation of Solution

Given data:

Charges, $\text{Q}\text{=}\text{4}\text{.0}\text{C}$

Formula used:

  $\text{Q}\text{=}\text{mcΔT}$

Where,

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

  $\text{m}$ is mass.

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

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

Calculation:

The energy (as the heat energy) is used to raise the temperature of the water and also, boil it.

Suppose the room temperature is $\text{20 °C}$ , then $\text{Q=}\text{mcΔT}\text{+}{\text{mL}}_{\text{f}}\mathrm{...}\left(\text{1}\right)$ .

Substitute the given values in equation $\left(1\right)$

  $\begin{array}{c}\text{m =}\frac{\text{Q}}{\text{c}\Delta {\text{T+L}}_{\text{f}}}\\ \text{=}\frac{\text{4}\text{.2}{\text{×10}}^{\text{6}}\text{J}}{\left(\text{4186}\frac{\text{J}}{\text{kg}\cdot °\text{C}}\right)\left(100°\text{C - }20°\text{C}\right)+22.6\times {10}^5\frac{\text{J}}{\text{kg}}}\\ =1.6\text{kg}\end{array}$