Q55PE - College Physics (Urone)

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?

Step-by-Step Solution

Answer

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

Step 1: Given information

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

The capacitance’s capacitor is:

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

Mass of human flesh: 

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

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

Step 2: 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.

Step 3: 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:

 $\mathrm{Q}=\mathrm{mc\Delta t}$

Step 4: 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-

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

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