Question

A prankster applies 450V to an$\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu F}$ capacitor and then tosses it to an unsuspecting victim. The victim's finger is burned by the discharge of the capacitor through 0.200 g of flesh. What is the temperature increase of the flesh? Is it reasonable to assume no phase change?

Short answer

The required solution is $\mathrm{\Delta T}=11.6°\mathrm{C}$

Step-by-step solution

Given information

Human’s body specific heat is: $\mathrm{c}=3500\mathrm{J}/\mathrm{kg}°\mathrm{C}$

The capacitance’s capacitor is:

$\begin{array}{rcl}\mathbf{C}& \mathbf{=}& \mathbf{(}\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu F}\mathbf{)}\left(\frac{1F}{{10}^6\mathrm{\mu F}}\right)\\ & \mathbf{=}& \mathbf{80}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathbf{F}\end{array}$

Mass of human flesh:

$\begin{array}{rcl}\mathbf{m}& \mathbf{=}& \mathbf{(}\mathbf{0}\mathbf{.}\mathbf{200}\mathbf{}\mathbf{g}\mathbf{)}\left(\frac{1\mathrm{kg}}{1000g}\right)\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{200}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{kg}\end{array}$

Across the capacitor, the potential difference is: $\mathrm{\Delta V}=450\mathrm{V}$

Defining capacitor

A two-terminal electrical component, which can store energy in the form of an electric field, is known as the capacitor and its ability to store energy is called capacitance.

Energy storage in a capacitor

The energy held in a capacitor that has a capacitance of C, a charge of q, and a potential difference $\mathrm{\Delta V}$is -

${\mathbf{U}}_{\mathbf{E}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{C}\mathbf{(}\mathbf{\Delta V}{\mathbf{)}}^{\mathbf{2}}$

The amount of energy Q needed to change the temperature of a mass m of a substance by $\mathrm{\Delta T}$and C is the heat of the substance:

$\mathrm{Q}=\mathrm{mc\Delta t}$

Observation of temperature raises

When the capacitor discharges and burns the victim's finger, the energy U(E) held in the capacitor is converted into the energy Q required to raise the flesh temperature:

$\mathbf{U}\left(E\right)\mathbf{=}\mathbf{Q}$

Substituting the values from other equations for U(E) and Q

$\frac{\mathbf{1}}{\mathbf{2}}\mathbf{C}\mathbf{(}\mathbf{\Delta V}{\mathbf{)}}^{\mathbf{2}}\mathbf{=}\mathbf{mc\Delta T}$

$\mathbf{\Delta T}\mathbf{=}\frac{\frac{\mathbf{1}}{\mathbf{2}}\mathbf{C}{\mathbf{\left(}\mathbf{\Delta V}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{mc}}$

Substitute the values-

$\begin{array}{rcl}\mathbf{∆}\mathbf{T}& \mathbf{=}& \frac{\left(80.0\times {10}^{-6}F\right){\left(450V\right)}^{\mathbf{2}}}{\mathbf{2}\left(0.200\times {10}^{-3}\mathrm{kg}\right)\left(3500J/\mathrm{kg}°C\right)}\\ & \mathbf{=}& \mathbf{11}\mathbf{.}\mathbf{6}\mathbf{°}\mathbf{C}\end{array}$

Therefore, the temperature raises is $\mathrm{\Delta T}=11.6°\mathrm{C}$